1.If the magnitude of the sum of two vectors is equal to the magnitude of the difference of the two vectors, what is the angle between the vectors?
Sol: Given $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$.
Squaring both sides: $A^2 + B^2 + 2AB\cos\theta = A^2 + B^2 - 2AB\cos\theta$.
$\implies 4AB\cos\theta = 0 \implies \cos\theta = 0 \implies \mathbf{\theta = 90^\circ}$.
2.Two forces $P$ and $Q$ of magnitude $2F$ and $3F$, respectively, are at an angle $\theta$ with each other. If the force $Q$ is doubled, then their resultant also gets doubled. Find the angle $\theta$.
Sol: Case 1: $R^2 = (2F)^2 + (3F)^2 + 2(2F)(3F)\cos\theta \implies R^2 = 13F^2 + 12F^2\cos\theta$.
Case 2: $Q$ is doubled to $6F$, $R$ doubles to $2R$.
$(2R)^2 = (2F)^2 + (6F)^2 + 2(2F)(6F)\cos\theta \implies 4R^2 = 40F^2 + 24F^2\cos\theta$.
Substitute $R^2$: $4(13F^2 + 12F^2\cos\theta) = 40F^2 + 24F^2\cos\theta$.
$52 + 48\cos\theta = 40 + 24\cos\theta \implies 24\cos\theta = -12 \implies \cos\theta = -1/2 \implies \mathbf{\theta = 120^\circ}$.
3.Given $\vec{A} = 3\hat{i} + 4\hat{j} + \hat{k}$ and $\vec{B} = \hat{i} - \hat{j} + \hat{k}$. Find the unit vector in the direction of $(\vec{A} + \vec{B})$.
Sol: $\vec{R} = \vec{A} + \vec{B} = (3+1)\hat{i} + (4-1)\hat{j} + (1+1)\hat{k} = 4\hat{i} + 3\hat{j} + 2\hat{k}$.
Magnitude $|\vec{R}| = \sqrt{4^2 + 3^2 + 2^2} = \sqrt{16 + 9 + 4} = \sqrt{29}$.
Unit vector $\hat{R} = \mathbf{\frac{4\hat{i} + 3\hat{j} + 2\hat{k}}{\sqrt{29}}}$.
4.Find the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{P} = 2\hat{i} + 3\hat{j}$ and $\vec{Q} = \hat{i} + 4\hat{j}$.
Sol: Area = $|\vec{P} \times \vec{Q}|$.
$\vec{P} \times \vec{Q} = (2\hat{i} + 3\hat{j}) \times (\hat{i} + 4\hat{j}) = (2)(4)(\hat{i}\times\hat{j}) + (3)(1)(\hat{j}\times\hat{i}) = 8\hat{k} - 3\hat{k} = 5\hat{k}$.
Area = $|5\hat{k}| = \mathbf{5 \text{ units}}$.
5.Find the component of vector $\vec{A} = 2\hat{i} + 3\hat{j}$ along the direction of vector $\vec{B} = \hat{i} + \hat{j}$.
Sol: Component of $\vec{A}$ along $\vec{B}$ is $\vec{A} \cdot \hat{B}$.
$\hat{B} = \frac{\hat{i} + \hat{j}}{\sqrt{1^2 + 1^2}} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.
$\vec{A} \cdot \hat{B} = (2\hat{i} + 3\hat{j}) \cdot \left(\frac{\hat{i} + \hat{j}}{\sqrt{2}}\right) = \frac{2(1) + 3(1)}{\sqrt{2}} = \mathbf{\frac{5}{\sqrt{2}}}$.
6.If vectors $\vec{P} = a\hat{i} + a\hat{j} + 3\hat{k}$ and $\vec{Q} = a\hat{i} - 2\hat{j} - \hat{k}$ are perpendicular to each other, find the positive value of $a$.
Sol: For perpendicular vectors, $\vec{P} \cdot \vec{Q} = 0$.
$(a)(a) + (a)(-2) + (3)(-1) = 0 \implies a^2 - 2a - 3 = 0$.
Factoring: $(a - 3)(a + 1) = 0 \implies a = 3 \text{ or } a = -1$.
Positive value is $\mathbf{a = 3}$.
7.A car travels due East with a speed of $10 \text{ m/s}$. It then takes a left turn and travels due North with the same speed of $10 \text{ m/s}$. Find the magnitude and direction of the change in velocity.
Sol: Initial velocity $\vec{v}_i = 10\hat{i}$. Final velocity $\vec{v}_f = 10\hat{j}$.
Change in velocity $\Delta\vec{v} = \vec{v}_f - \vec{v}_i = 10\hat{j} - 10\hat{i}$.
Magnitude $|\Delta\vec{v}| = \sqrt{(-10)^2 + (10)^2} = \sqrt{200} = \mathbf{10\sqrt{2} \text{ m/s}}$.
Direction: $-x$ and $+y$ implies North-West (NW). Angle is $135^\circ$ from East.
8.If $\vec{A} \times \vec{B} = \vec{B} \times \vec{A}$, what can you conclude about the angle between vectors $\vec{A}$ and $\vec{B}$?
Sol: We know $\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})$.
Given $\vec{A} \times \vec{B} = \vec{B} \times \vec{A} \implies \vec{A} \times \vec{B} = -(\vec{A} \times \vec{B}) \implies 2(\vec{A} \times \vec{B}) = 0 \implies \vec{A} \times \vec{B} = \vec{0}$.
This implies $AB\sin\theta\hat{n} = \vec{0} \implies \sin\theta = 0$.
Thus, the angle $\theta$ is $\mathbf{0^\circ \text{ or } 180^\circ}$ (vectors are parallel or anti-parallel).
9.The sum of the magnitudes of two forces acting at a point is $18 \text{ N}$ and the magnitude of their resultant is $12 \text{ N}$. If the resultant is at $90^\circ$ with the force of smaller magnitude, what are the magnitudes of forces?
Sol: Let forces be $P$ and $Q$, with $P < Q$. Given $P + Q = 18$.
Resultant $\vec{R} \perp \vec{P} \implies$ vector triangle forms a right triangle with $Q$ as hypotenuse, $P$ as base, $R$ as perpendicular.
$Q^2 = P^2 + R^2 \implies Q^2 - P^2 = R^2 \implies (Q-P)(Q+P) = 12^2 = 144$.
Substitute $Q+P = 18 \implies (Q-P)(18) = 144 \implies Q-P = 8$.
Solving $Q+P=18$ and $Q-P=8$: Add them $\implies 2Q = 26 \implies \mathbf{Q = 13 \text{ N}}$. Then $\mathbf{P = 5 \text{ N}}$.
10.Prove that the vectors $\vec{A} = 2\hat{i} - 3\hat{j} + \hat{k}$ and $\vec{B} = 4\hat{i} - 6\hat{j} + 2\hat{k}$ are parallel to each other.
Sol: Two vectors are parallel if one is a scalar multiple of the other.
$\vec{B} = 4\hat{i} - 6\hat{j} + 2\hat{k} = 2(2\hat{i} - 3\hat{j} + \hat{k}) = 2\vec{A}$.
Since $\vec{B} = 2\vec{A}$, they have the exact same direction. (Proved).
11.The position of a particle is given by $\vec{r} = 3t\hat{i} + (2t^2 - 5t)\hat{j} + 4\hat{k} \text{ m}$. Find the magnitude of its velocity and acceleration at $t = 2 \text{ s}$.
Sol: $\vec{v} = \frac{d\vec{r}}{dt} = 3\hat{i} + (4t - 5)\hat{j}$.
At $t=2\text{ s}$, $\vec{v} = 3\hat{i} + (8-5)\hat{j} = 3\hat{i} + 3\hat{j} \implies |\vec{v}| = \sqrt{3^2 + 3^2} = \mathbf{3\sqrt{2} \text{ m/s}}$.
$\vec{a} = \frac{d\vec{v}}{dt} = 4\hat{j}$. Magnitude $|\vec{a}| = \mathbf{4 \text{ m/s}^2}$.
(Note: Original placeholder key stated v=5; mathematically $3\sqrt{2} \approx 4.24$ is exact based on the given function).
12.A particle moves in the $x$-$y$ plane with a constant acceleration of $1.5 \text{ m/s}^2$ in the direction making an angle of $37^\circ$ with the x-axis. At $t=0$, the particle is at the origin and its velocity is $8.0 \text{ m/s}$ along the x-axis. Find the velocity and the position of the particle at $t = 4.0 \text{ s}$.
Sol: $\vec{a} = 1.5\cos 37^\circ \hat{i} + 1.5\sin 37^\circ \hat{j} = 1.5(0.8)\hat{i} + 1.5(0.6)\hat{j} = 1.2\hat{i} + 0.9\hat{j} \text{ m/s}^2$.
$\vec{u} = 8.0\hat{i}$.
$\vec{v}(t) = \vec{u} + \vec{a}t = 8.0\hat{i} + (1.2\hat{i} + 0.9\hat{j})(4) = 8.0\hat{i} + 4.8\hat{i} + 3.6\hat{j} = \mathbf{12.8\hat{i} + 3.6\hat{j} \text{ m/s}}$.
$\vec{r}(t) = \vec{u}t + \frac{1}{2}\vec{a}t^2 = 8(4)\hat{i} + \frac{1}{2}(1.2\hat{i} + 0.9\hat{j})(16) = 32\hat{i} + 9.6\hat{i} + 7.2\hat{j} = \mathbf{41.6\hat{i} + 7.2\hat{j} \text{ m}}$.
13.The coordinates of a moving particle at any time $t$ are given by $x = \alpha t^3$ and $y = \beta t^3$. Find the speed of the particle at time $t$.
Sol: $v_x = \frac{dx}{dt} = 3\alpha t^2$. $v_y = \frac{dy}{dt} = 3\beta t^2$.
Speed $|\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{(3\alpha t^2)^2 + (3\beta t^2)^2} = \sqrt{9t^4(\alpha^2 + \beta^2)} = \mathbf{3t^2\sqrt{\alpha^2 + \beta^2}}$.
14.A particle starts from origin at $t=0$ with a velocity $5.0\hat{i} \text{ m/s}$ and moves in $x$-$y$ plane under action of a force which produces a constant acceleration of $(3.0\hat{i} + 2.0\hat{j}) \text{ m/s}^2$. What is the y-coordinate of the particle at the instant its x-coordinate is $84 \text{ m}$?
Sol: x-motion: $x = u_x t + \frac{1}{2}a_x t^2 \implies 84 = 5t + \frac{1}{2}(3)t^2 \implies 1.5t^2 + 5t - 84 = 0$.
Multiply by 2: $3t^2 + 10t - 168 = 0$. Solving for $t$: $t = \frac{-10 \pm \sqrt{100 - 4(3)(-168)}}{6} = \frac{-10 \pm 46}{6}$.
Positive time $t = 36/6 = 6\text{ s}$.
y-motion: $y = u_y t + \frac{1}{2}a_y t^2 = 0 + \frac{1}{2}(2)(6^2) = 1(36) = \mathbf{36 \text{ m}}$.
15.The equation of trajectory of a particle is given by $y = 2x - 5x^2$ (where $x, y$ are in meters). If the horizontal component of velocity is constant at $10 \text{ m/s}$, find the initial velocity of the particle and its acceleration.
Sol: $v_x = \frac{dx}{dt} = 10\text{ m/s}$.
Differentiate trajectory wrt time: $v_y = \frac{dy}{dt} = 2\frac{dx}{dt} - 10x\frac{dx}{dt} = 2(10) - 10x(10) = 20 - 100x$.
At origin ($x=0$), $v_{y0} = 20$. Initial velocity $\vec{u} = \mathbf{10\hat{i} + 20\hat{j} \text{ m/s}}$.
Acceleration $\vec{a} = a_x\hat{i} + a_y\hat{j}$. Since $v_x$ is constant, $a_x = 0$.
$a_y = \frac{dv_y}{dt} = \frac{d}{dt}(20 - 100x) = -100\frac{dx}{dt} = -100(10) = -1000\text{ m/s}^2$.
$\vec{a} = \mathbf{-1000\hat{j} \text{ m/s}^2}$. (Correction from short key: proper differentiation yields 1000).
16.If a particle moves in a 2D plane such that its velocity components are $v_x = 4 \text{ m/s}$ and $v_y = (6x - 4) \text{ m/s}$, find the equation of its path if it starts from the origin.
Sol: $v_x = \frac{dx}{dt} = 4 \implies dx = 4 dt \implies x = 4t$ (since starts at origin).
$v_y = \frac{dy}{dt} = 6x - 4$. Since we need $y=f(x)$, apply chain rule: $\frac{dy}{dx} = \frac{v_y}{v_x} = \frac{6x - 4}{4} = \frac{3}{2}x - 1$.
Integrate wrt $x$: $\int dy = \int (\frac{3}{2}x - 1) dx \implies y = \frac{3}{4}x^2 - x + C$.
At $x=0, y=0 \implies C=0$. Path equation: $\mathbf{y = \frac{3x^2}{4} - x}$.
17.The position vector of a particle is $\vec{r} = (A \cos\omega t)\hat{i} + (A \sin\omega t)\hat{j}$. Prove that the velocity vector is perpendicular to the position vector.
Sol: Velocity $\vec{v} = \frac{d\vec{r}}{dt} = (-A\omega\sin\omega t)\hat{i} + (A\omega\cos\omega t)\hat{j}$.
Take dot product $\vec{r} \cdot \vec{v} = (A\cos\omega t)(-A\omega\sin\omega t) + (A\sin\omega t)(A\omega\cos\omega t)$.
$= -A^2\omega\sin\omega t\cos\omega t + A^2\omega\sin\omega t\cos\omega t = 0$.
Since $\vec{r} \cdot \vec{v} = 0$, the velocity vector is strictly perpendicular to the position vector. (Proved).
18.A projectile is fired with a velocity of $30 \text{ m/s}$ at an angle of $60^\circ$ with the horizontal. Calculate its (i) time of flight, (ii) maximum height, and (iii) horizontal range. ($g = 10 \text{ m/s}^2$)
Sol: $u = 30$, $\theta = 60^\circ$.
(i) $T = \frac{2u\sin\theta}{g} = \frac{2(30)\sin 60^\circ}{10} = 6 \left(\frac{\sqrt{3}}{2}\right) = \mathbf{3\sqrt{3} \text{ s}}$.
(ii) $H = \frac{u^2\sin^2\theta}{2g} = \frac{30^2(3/4)}{20} = \frac{900 \times 0.75}{20} = \frac{675}{20} = \mathbf{33.75 \text{ m}}$.
(iii) $R = \frac{u^2\sin 2\theta}{g} = \frac{900\sin 120^\circ}{10} = 90 \left(\frac{\sqrt{3}}{2}\right) = \mathbf{45\sqrt{3} \text{ m}}$.
19.For a projectile, the horizontal range is equal to its maximum height. What is the angle of projection?
Sol: $R = H \implies \frac{u^2\sin 2\theta}{g} = \frac{u^2\sin^2\theta}{2g}$.
$2\sin\theta\cos\theta = \frac{\sin^2\theta}{2}$. Since $\theta \ne 0$, divide by $\sin\theta$:
$2\cos\theta = \frac{\sin\theta}{2} \implies 4 = \frac{\sin\theta}{\cos\theta} \implies \tan\theta = 4 \implies \mathbf{\theta = \tan^{-1}(4)}$.
20.Show that there are two angles of projection for which the horizontal range is the same. If $t_1$ and $t_2$ are the times of flight for these two angles, prove that $t_1 t_2 = 2R/g$.
Sol: $R = \frac{u^2\sin 2\theta}{g}$. Since $\sin(180^\circ - 2\theta) = \sin 2\theta$, angle $\theta' = 90^\circ - \theta$ yields the same range. Thus, $\theta$ and $90^\circ-\theta$ are complementary angles.
$t_1 = \frac{2u\sin\theta}{g}$ and $t_2 = \frac{2u\sin(90^\circ-\theta)}{g} = \frac{2u\cos\theta}{g}$.
Product $t_1 t_2 = \left(\frac{2u\sin\theta}{g}\right)\left(\frac{2u\cos\theta}{g}\right) = \frac{2}{g} \left( \frac{2u^2\sin\theta\cos\theta}{g} \right) = \frac{2}{g} \left( \frac{u^2\sin 2\theta}{g} \right) = \mathbf{\frac{2R}{g}}$. (Proved).
21.The maximum horizontal range of a projectile is $400 \text{ m}$. What is the maximum height attained by it in this case?
Sol: Max range occurs at $\theta = 45^\circ$.
$R_{max} = \frac{u^2}{g} = 400 \implies u^2 = 400g$.
Height at this angle: $H = \frac{u^2\sin^2 45^\circ}{2g} = \frac{u^2 (1/2)}{2g} = \frac{u^2}{4g}$.
Substitute $u^2 = 400g$: $H = \frac{400g}{4g} = \mathbf{100 \text{ m}}$.
22.A boy can throw a stone up to a maximum height of $10 \text{ m}$. What is the maximum horizontal distance to which he can throw the same stone?
Sol: Max vertical throw is at $\theta = 90^\circ$. $H_{max} = \frac{u^2}{2g} = 10 \implies \frac{u^2}{g} = 20$.
Max horizontal range is at $\theta = 45^\circ$. $R_{max} = \frac{u^2}{g}$.
Thus, $R_{max} = \mathbf{20 \text{ m}}$.
23.A projectile is given an initial velocity of $(\hat{i} + 2\hat{j}) \text{ m/s}$. Find the equation of its trajectory. ($g=10 \text{ m/s}^2$)
Sol: $u_x = 1 \text{ m/s}, u_y = 2 \text{ m/s}$. $\tan\theta = \frac{u_y}{u_x} = 2$.
Equation of trajectory: $y = x\tan\theta - \frac{gx^2}{2u_x^2}$.
$y = x(2) - \frac{10x^2}{2(1^2)} = \mathbf{2x - 5x^2}$.
24.A body is projected at such an angle that the horizontal range is $3$ times the maximum height. Find the angle of projection.
Sol: $R = 3H \implies \frac{u^2\sin 2\theta}{g} = 3 \frac{u^2\sin^2\theta}{2g}$.
$2\sin\theta\cos\theta = 1.5\sin^2\theta \implies \frac{\sin\theta}{\cos\theta} = \frac{2}{1.5} = \frac{4}{3}$.
$\tan\theta = \frac{4}{3} \implies \mathbf{\theta = 53^\circ}$.
25.A bullet is fired at an angle of $30^\circ$ with the horizontal and hits the ground $3.0 \text{ km}$ away. By adjusting its angle of projection, can one hope to hit a target $5.0 \text{ km}$ away? Assume the muzzle speed to be fixed, and neglect air resistance.
Sol: Current range $R = \frac{u^2\sin 60^\circ}{g} = 3.0 \text{ km}$.
$\frac{u^2}{g} (\frac{\sqrt{3}}{2}) = 3 \implies \frac{u^2}{g} = \frac{6}{\sqrt{3}} = 2\sqrt{3} \approx 3.46 \text{ km}$.
The absolute maximum range ($R_{max} = u^2/g$ at $45^\circ$) is $3.46 \text{ km}$.
Since $5.0 \text{ km} > 3.46 \text{ km}$, No, it cannot hit the target.
26.At what time after projection will the velocity vector of a projectile, thrown with speed $u$ at an angle $\theta$, become perpendicular to its initial velocity vector?
Sol: Initial velocity $\vec{u} = u\cos\theta\hat{i} + u\sin\theta\hat{j}$.
Velocity at time $t$: $\vec{v} = u\cos\theta\hat{i} + (u\sin\theta - gt)\hat{j}$.
For $\vec{v} \perp \vec{u}$, dot product $\vec{u} \cdot \vec{v} = 0$.
$(u\cos\theta)(u\cos\theta) + (u\sin\theta)(u\sin\theta - gt) = 0 \implies u^2\cos^2\theta + u^2\sin^2\theta - ugt\sin\theta = 0$.
$u^2(\cos^2\theta + \sin^2\theta) = ugt\sin\theta \implies u^2 = ugt\sin\theta \implies \mathbf{t = \frac{u}{g\sin\theta}}$.
27.A particle is projected with a speed $v_0$ at an angle $\theta$ with the horizontal. Find the average velocity of the projectile between the point of projection and the highest point of its trajectory.
Sol: Coordinate of highest point: $x = R/2 = \frac{v_0^2\sin\theta\cos\theta}{g}$, $y = H = \frac{v_0^2\sin^2\theta}{2g}$. Time $t = \frac{v_0\sin\theta}{g}$.
Average velocity vector $\vec{v}_{avg} = \frac{\Delta x}{t}\hat{i} + \frac{\Delta y}{t}\hat{j}$.
$v_{avg,x} = \frac{(v_0^2\sin\theta\cos\theta)/g}{v_0\sin\theta/g} = v_0\cos\theta$.
$v_{avg,y} = \frac{(v_0^2\sin^2\theta)/2g}{v_0\sin\theta/g} = \frac{v_0\sin\theta}{2}$.
Magnitude $|\vec{v}_{avg}| = \sqrt{(v_0\cos\theta)^2 + \left(\frac{v_0\sin\theta}{2}\right)^2} = \frac{v_0}{2}\sqrt{4\cos^2\theta + \sin^2\theta} = \mathbf{\frac{v_0}{2}\sqrt{1 + 3\cos^2\theta}}$.
28.The equation of a projectile's path is $y = \sqrt{3}x - \frac{gx^2}{2}$. Determine the angle of projection and initial velocity.
Sol: Compare with standard equation: $y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$.
$\tan\theta = \sqrt{3} \implies \mathbf{\theta = 60^\circ}$.
$\frac{g}{2} = \frac{g}{2u^2\cos^2 60^\circ} \implies 1 = \frac{1}{u^2 (1/4)} \implies u^2 \left(\frac{1}{4}\right) = 1 \implies u^2 = 4 \implies \mathbf{u = 2 \text{ m/s}}$.
29.A particle is projected from the ground with an initial speed of $u$ at an angle $\theta$ with the horizontal. What is the change in the magnitude of its momentum when it strikes the ground? What is the change in its momentum vector?
Sol: Initial speed is $u$. It strikes ground with same speed $u$. Magnitude of momentum $|p| = mu$. Change in magnitude is $\mathbf{0}$.
Initial momentum vector $\vec{p}_i = mu\cos\theta\hat{i} + mu\sin\theta\hat{j}$.
Final momentum vector $\vec{p}_f = mu\cos\theta\hat{i} - mu\sin\theta\hat{j}$.
Change $\Delta\vec{p} = \vec{p}_f - \vec{p}_i = \mathbf{-2mu\sin\theta\hat{j}}$.
30.Two particles are projected from the same point with the same speed $u$ such that they have the same range $R$, but different maximum heights $h_1$ and $h_2$. Prove that $R = 4\sqrt{h_1 h_2}$.
Sol: Same range implies angles are $\theta$ and $90^\circ-\theta$.
$h_1 = \frac{u^2\sin^2\theta}{2g}$, $h_2 = \frac{u^2\sin^2(90^\circ-\theta)}{2g} = \frac{u^2\cos^2\theta}{2g}$.
Evaluate $4\sqrt{h_1 h_2} = 4\sqrt{\left(\frac{u^2\sin^2\theta}{2g}\right)\left(\frac{u^2\cos^2\theta}{2g}\right)} = 4\sqrt{\frac{u^4\sin^2\theta\cos^2\theta}{4g^2}} = 4 \left( \frac{u^2\sin\theta\cos\theta}{2g} \right)$.
$= \frac{2u^2\sin\theta\cos\theta}{g} = \frac{u^2\sin 2\theta}{g} = \mathbf{R}$. (Proved).
31.A ball whose kinetic energy is $E$ is projected at an angle of $45^\circ$ to the horizontal. What is the kinetic energy of the ball at the highest point of its flight?
Sol: Initial KE $E = \frac{1}{2}mu^2$.
At highest point, vertical velocity is $0$, horizontal is $u_x = u\cos 45^\circ = u/\sqrt{2}$.
KE at highest point $K' = \frac{1}{2}mv_{top}^2 = \frac{1}{2}m(u/\sqrt{2})^2 = \frac{1}{2}m(u^2/2) = \frac{1}{2}\left(\frac{1}{2}mu^2\right) = \mathbf{\frac{E}{2}}$.
32.A bomber plane flying horizontally with a speed of $500 \text{ m/s}$ at a height of $1000 \text{ m}$ drops a bomb. Find the time taken by the bomb to hit the ground and the horizontal distance it travels. ($g = 10 \text{ m/s}^2$)
Sol: Vertical motion is free fall: $t = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2(1000)}{10}} = \sqrt{200} = 10\sqrt{2} \approx \mathbf{14.14 \text{ s}}$.
Horizontal distance $R = u_x t = 500 \times 10\sqrt{2} = 5000\sqrt{2} \approx \mathbf{7070 \text{ m}}$.
33.A ball is thrown horizontally from the top of a tower $40 \text{ m}$ high. The ball strikes the ground at a point $80 \text{ m}$ from the foot of the tower. Find the angle that the velocity vector makes with the horizontal just before hitting the ground. ($g = 10 \text{ m/s}^2$)
Sol: Time of flight $t = \sqrt{2H/g} = \sqrt{2(40)/10} = \sqrt{8} = 2\sqrt{2}\text{ s}$.
Horizontal velocity $u_x = R/t = 80/(2\sqrt{2}) = 20\sqrt{2}\text{ m/s}$.
Vertical velocity at impact $v_y = gt = 10(2\sqrt{2}) = 20\sqrt{2}\text{ m/s}$.
Angle with horizontal $\tan\theta = \frac{v_y}{v_x} = \frac{20\sqrt{2}}{20\sqrt{2}} = 1 \implies \mathbf{\theta = 45^\circ}$.
34.A stone is thrown upwards from the top of a $20 \text{ m}$ high building with an initial velocity of $10 \text{ m/s}$ at an angle of $30^\circ$ above the horizontal. Find the horizontal range from the foot of the building. ($g = 10 \text{ m/s}^2$)
Sol: $u_x = 10\cos 30^\circ = 5\sqrt{3}$, $u_y = 10\sin 30^\circ = 5$. Displacement $S_y = -20\text{ m}$.
$S_y = u_y t - \frac{1}{2}gt^2 \implies -20 = 5t - 5t^2 \implies t^2 - t - 4 = 0$.
$t = \frac{1 \pm \sqrt{1 - 4(1)(-4)}}{2} = \frac{1 + \sqrt{17}}{2} \approx \frac{1 + 4.123}{2} \approx 2.56\text{ s}$.
Range $R = u_x t = 5\sqrt{3} \times 2.56 \approx 8.66 \times 2.56 \approx \mathbf{22.2 \text{ m}}$. (Exact algebraic: $5\sqrt{3}(1+\sqrt{17})/2$).
35.Two tall buildings are separated by a distance of $50 \text{ m}$. A person throws a ball horizontally from a window $45 \text{ m}$ above the ground in the first building. With what speed must the ball be thrown so that it enters a window $25 \text{ m}$ above the ground in the second building? ($g = 10 \text{ m/s}^2$)
Sol: Vertical drop $h = 45 - 25 = 20\text{ m}$.
Time to drop $20\text{ m}$: $t = \sqrt{2h/g} = \sqrt{2(20)/10} = \sqrt{4} = 2\text{ s}$.
Horizontal distance $R = 50\text{ m}$.
Speed $u = R/t = 50 / 2 = \mathbf{25 \text{ m/s}}$.
36.A particle is projected up an inclined plane of inclination $\beta = 30^\circ$ with an initial velocity $u = 20 \text{ m/s}$ at an angle $\alpha = 60^\circ$ with the horizontal. Find the range of the projectile along the inclined plane. ($g = 10 \text{ m/s}^2$)
Sol: Angle of projection with incline $\theta = \alpha - \beta = 60^\circ - 30^\circ = 30^\circ$.
Range on incline $R = \frac{2u^2\sin\theta\cos(\theta+\beta)}{g\cos^2\beta} = \frac{2(20^2)\sin 30^\circ\cos 60^\circ}{10\cos^2 30^\circ}$.
$R = \frac{2(400)(1/2)(1/2)}{10(3/4)} = \frac{200}{7.5} = \frac{200 \times 4}{30} = \frac{80}{3} \approx \mathbf{26.67 \text{ m}}$.
37.A projectile is fired up an inclined plane (angle of inclination $\theta$). The angle of projection is $\alpha$ with the inclined plane. For the projectile to hit the inclined plane perpendicularly, what must be the relation between $\alpha$ and $\theta$?
Sol: Let x-axis be along incline. $u_x = u\cos\alpha$, $a_x = -g\sin\theta$.
For perpendicular hit, $v_x = 0$ at time of flight $T$.
$v_x = u_x + a_x T \implies 0 = u\cos\alpha - (g\sin\theta)T \implies T = \frac{u\cos\alpha}{g\sin\theta}$.
Standard time of flight $T = \frac{2u_y}{|a_y|} = \frac{2u\sin\alpha}{g\cos\theta}$.
Equating: $\frac{u\cos\alpha}{g\sin\theta} = \frac{2u\sin\alpha}{g\cos\theta} \implies \frac{\cos\alpha}{\sin\alpha} = 2\frac{\sin\theta}{\cos\theta} \implies \mathbf{\cot\alpha = 2\tan\theta}$.
38.A truck is moving with a constant velocity of $10 \text{ m/s}$. A boy inside the truck throws a ball vertically upwards with a speed of $20 \text{ m/s}$ relative to the truck. Where will the ball fall? What is the trajectory seen by an observer on the ground?
Sol: Relative to the truck, the ball's horizontal velocity is 0, so it will fall straight back into the boy's hands.
Relative to an observer on the ground, the ball has horizontal velocity $u_x = 10\text{ m/s}$ (truck's speed) and vertical $u_y = 20\text{ m/s}$. The trajectory observed will be a Parabola.
39.Two particles are projected simultaneously from the same point, with the same speed $u$, but at different angles $\theta_1$ and $\theta_2$ such that their ranges are equal. Prove that the distance between them is always purely vertical and find the separation as a function of time.
Sol: *Note: A known physical flaw exists in this classic phrasing. If projected with same $u$ and equal ranges, $\theta_1 = 90^\circ-\theta_2$. Their horizontal velocities $u\cos\theta_1 \ne u\cos\theta_2$ are different, so separation is NEVER purely vertical.*
If the question implied equal *horizontal velocities* (different $u$), then $x_1(t) = x_2(t)$.
Taking the literal vertical separation $\Delta y$: $y_1 = u\sin\theta_1 t - \frac{1}{2}gt^2$ and $y_2 = u\sin\theta_2 t - \frac{1}{2}gt^2$.
$\Delta y = y_1 - y_2 = \mathbf{ut(\sin\theta_1 - \sin\theta_2)}$.
40.From the top of a tower, two stones are projected simultaneously with the same speed $u$; one upwards at angle $\theta$ and the other downwards at the same angle $\theta$ with the horizontal. Which stone will strike the ground with greater speed?
Sol: By conservation of energy: Total Energy at top $= \frac{1}{2}mu^2 + mgH$.
This energy must equal the kinetic energy at the ground: $\frac{1}{2}mv^2$.
Thus, $v = \sqrt{u^2 + 2gH}$. Since both have the same initial speed $u$ and height $H$, they will hit the ground with the exact same speed (though at different angles and times).
41.Rain is falling vertically with a speed of $30 \text{ m/s}$. A woman rides a bicycle with a speed of $10 \text{ m/s}$ in the north to south direction. What is the direction in which she should hold her umbrella?
Sol: $\vec{v}_r = -30\hat{j}$ (down). $\vec{v}_w = -10\hat{i}$ (South, assuming North is $+x$).
Velocity of rain wrt woman $\vec{v}_{rw} = \vec{v}_r - \vec{v}_w = -30\hat{j} - (-10\hat{i}) = 10\hat{i} - 30\hat{j}$.
Angle with vertical $\tan\theta = \frac{|v_x|}{|v_y|} = \frac{10}{30} = \frac{1}{3}$.
Umbrella must be held $\mathbf{\theta = \tan^{-1}(1/3)}$ South of vertical.
42.A man walking horizontally at $3 \text{ km/h}$ observes rain falling vertically. When he increases his speed to $6 \text{ km/h}$, the rain appears to hit him at an angle of $45^\circ$ with the vertical. Find the actual speed and direction of the rain.
Sol: Let actual rain velocity be $\vec{v}_r = v_x\hat{i} - v_y\hat{j}$.
Case 1: Man velocity $\vec{v}_{m1} = 3\hat{i}$. $\vec{v}_{r/m1} = (v_x-3)\hat{i} - v_y\hat{j}$. Observed vertical implies $v_x - 3 = 0 \implies v_x = 3$.
Case 2: Man velocity $\vec{v}_{m2} = 6\hat{i}$. $\vec{v}_{r/m2} = (3-6)\hat{i} - v_y\hat{j} = -3\hat{i} - v_y\hat{j}$.
Angle $45^\circ \implies \tan 45^\circ = \frac{|v_x|}{|v_y|} \implies 1 = \frac{3}{v_y} \implies v_y = 3$.
Actual $\vec{v}_r = 3\hat{i} - 3\hat{j}$. Speed $|\vec{v}_r| = \sqrt{3^2 + 3^2} = \mathbf{3\sqrt{2} \text{ km/h}}$ at angle $\mathbf{45^\circ}$ with vertical.
43.A river is flowing from west to east at a speed of $5 \text{ m/s}$. A man on the south bank of the river, capable of swimming at $10 \text{ m/s}$ in still water, wants to swim across the river in the shortest time. He should swim in which direction? Also find the shortest time if the river width is $100 \text{ m}$.
Sol: For shortest time, all swimming effort must be perpendicular to the flow.
Swim direction: Due North ($90^\circ$ to river flow).
Time $t = \frac{\text{width}}{v_s} = \frac{100}{10} = \mathbf{10 \text{ s}}$.
44.In the previous question, if the man wants to cross the river to reach exactly the opposite point on the north bank (shortest path), at what angle with the river flow should he swim? How much time will it take?
Sol: To have zero drift, $v_s\sin\theta = v_r$ (where $\theta$ is angle upstream from vertical).
$10\sin\theta = 5 \implies \sin\theta = 1/2 \implies \theta = 30^\circ$ West of North.
Angle with flow (East) = $90^\circ + 30^\circ = \mathbf{120^\circ}$.
Net crossing speed $v_{net} = v_s\cos\theta = 10\cos 30^\circ = 5\sqrt{3} \text{ m/s}$.
Time $t = \frac{100}{5\sqrt{3}} = \frac{20}{\sqrt{3}} \approx \mathbf{11.55 \text{ s}}$.
45.A river flows with a speed $v_r$. A swimmer can swim with speed $v_s$ in still water. If $v_r > v_s$, can the swimmer cross the river strictly along the shortest path? If not, what should be his angle to minimize the drift?
Sol: To cross exactly opposite, requires $v_s\sin\theta = v_r \implies \sin\theta = v_r/v_s$. Since $v_r > v_s$, $\sin\theta > 1$ which is impossible. No, he cannot.
To minimize drift $x = (v_r - v_s\sin\theta) \frac{d}{v_s\cos\theta}$. Differentiating $dx/d\theta = 0$ yields condition $\mathbf{\sin\theta = v_s/v_r}$ upstream from the perpendicular.
46.Ship A is sailing towards north at $10\sqrt{2} \text{ km/h}$. Ship B, located $100 \text{ km}$ East of A, is sailing towards west at $10\sqrt{2} \text{ km/h}$. Find the minimum distance between the ships and the time when it occurs.
Sol: Let A start at $(0,0)$. $\vec{v}_A = 10\sqrt{2}\hat{j}$.
B starts at $(100,0)$. $\vec{v}_B = -10\sqrt{2}\hat{i}$.
Relative velocity $\vec{v}_{BA} = \vec{v}_B - \vec{v}_A = -10\sqrt{2}\hat{i} - 10\sqrt{2}\hat{j}$. $|\vec{v}_{BA}| = \sqrt{200+200} = 20\text{ km/h}$.
Initial relative pos $\vec{r}_{BA,0} = 100\hat{i}$.
Time $t = \frac{-\vec{r} \cdot \vec{v}}{v^2} = \frac{-(100\hat{i}) \cdot (-10\sqrt{2}\hat{i} - 10\sqrt{2}\hat{j})}{400} = \frac{1000\sqrt{2}}{400} = \mathbf{2.5\sqrt{2} \text{ hours}}$ (approx $3.53\text{h}$).
Min distance $d = \frac{|\vec{r} \times \vec{v}|}{|\vec{v}|} = \frac{|100\hat{i} \times (-10\sqrt{2}\hat{i} - 10\sqrt{2}\hat{j})|}{20} = \frac{|-1000\sqrt{2}\hat{k}|}{20} = \mathbf{50\sqrt{2} \text{ km}}$.
47.An airplane is flying with a velocity of $800 \text{ km/h}$ relative to the air. A wind is blowing from the south at $150 \text{ km/h}$. If the pilot wants to travel due east, what direction must he head his plane?
Sol: Wind from South means wind blows towards North: $\vec{v}_w = 150\hat{j}$.
Plane wants resultant $\vec{v}_p$ along $+\hat{i}$ (East). $\vec{v}_p = \vec{v}_{p/w} + \vec{v}_w$.
Vertical component must be zero: $v_{p/w}\sin\theta + 150 = 0 \implies 800\sin\theta = -150 \implies \sin\theta = -\frac{150}{800} = -\frac{3}{16}$.
Direction must be $\mathbf{\sin^{-1}(3/16)}$ South of East.
48.Two particles A and B start from origin and move on x and y axes with constant velocities $v_1$ and $v_2$ respectively. Find the rate of separation between them as a function of time.
Sol: Position of A: $\vec{r}_A = v_1 t \hat{i}$. Position of B: $\vec{r}_B = v_2 t \hat{j}$.
Distance $s = |\vec{r}_B - \vec{r}_A| = \sqrt{(-v_1 t)^2 + (v_2 t)^2} = t\sqrt{v_1^2 + v_2^2}$.
Rate of separation $\frac{ds}{dt} = \mathbf{\sqrt{v_1^2 + v_2^2}}$.
49.A person moving at $4 \text{ km/h}$ finds rain falling vertically. If he stops, the rain appears to fall at an angle of $30^\circ$ with the vertical. What is the actual velocity of the rain?
Sol: Let $\vec{v}_r = v_x\hat{i} - v_y\hat{j}$. Person $\vec{v}_m = 4\hat{i}$.
When moving: $\vec{v}_{rm} = (v_x - 4)\hat{i} - v_y\hat{j}$. Vertical means $x$-component is 0 $\implies v_x = 4$.
When stopped: $\vec{v}_r = 4\hat{i} - v_y\hat{j}$. Angle $30^\circ$ with vertical $\implies \tan 30^\circ = \frac{|v_x|}{|v_y|} = \frac{4}{v_y} \implies \frac{1}{\sqrt{3}} = \frac{4}{v_y} \implies v_y = 4\sqrt{3}$.
Magnitude $|\vec{v}_r| = \sqrt{4^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = \mathbf{8 \text{ km/h}}$. Angle is $30^\circ$.
50.A particle moves in a circle of radius $20 \text{ cm}$ with a linear speed of $10 \text{ m/s}$. Find the angular velocity of the particle.
Sol: Radius $r = 20\text{ cm} = 0.2\text{ m}$. Linear speed $v = 10\text{ m/s}$.
Relation $v = \omega r \implies \omega = \frac{v}{r} = \frac{10}{0.2} = \mathbf{50 \text{ rad/s}}$.
51.The angular position of a particle revolving on a circular path is given by $\theta(t) = 2t^3 - 4t^2 + 6t$. Find its angular acceleration at $t = 2 \text{ s}$.
Sol: Angular velocity $\omega = \frac{d\theta}{dt} = 6t^2 - 8t + 6$.
Angular acceleration $\alpha = \frac{d\omega}{dt} = 12t - 8$.
At $t = 2\text{ s}$, $\alpha = 12(2) - 8 = 24 - 8 = \mathbf{16 \text{ rad/s}^2}$.
52.A particle is moving in a circle of radius $R$ with constant speed $v$. What is the magnitude of change in its velocity after it covers half of the circle? What is the average acceleration during this time?
Sol: After half circle, velocity reverses direction (e.g., from $+v\hat{j}$ to $-v\hat{j}$).
$|\Delta\vec{v}| = |(-v\hat{j}) - (v\hat{j})| = |-2v\hat{j}| = \mathbf{2v}$.
Time taken $t = \frac{\text{Distance}}{\text{Speed}} = \frac{\pi R}{v}$.
Avg acceleration $|\vec{a}_{avg}| = \frac{|\Delta\vec{v}|}{t} = \frac{2v}{\pi R / v} = \mathbf{\frac{2v^2}{\pi R}}$.
53.An object moves in a circular path of radius $R = 2.5 \text{ m}$ such that its speed varies with time as $v = 2t$. Find the net acceleration of the object at $t = 1 \text{ s}$.
Sol: Tangential acceleration $a_t = \frac{dv}{dt} = 2 \text{ m/s}^2$.
At $t=1$, $v = 2(1) = 2 \text{ m/s}$.
Centripetal acceleration $a_c = \frac{v^2}{R} = \frac{2^2}{2.5} = \frac{4}{2.5} = 1.6 \text{ m/s}^2$.
Net acceleration $a = \sqrt{a_t^2 + a_c^2} = \sqrt{2^2 + 1.6^2} = \sqrt{4 + 2.56} = \sqrt{6.56} \approx \mathbf{2.56 \text{ m/s}^2}$.
54.In non-uniform circular motion, what is the angle between the velocity vector and the net acceleration vector if the particle is speeding up? What if it is slowing down?
Sol: Net acceleration $\vec{a} = \vec{a}_c + \vec{a}_t$. Velocity $\vec{v}$ is along the tangent.
Speeding up: $\vec{a}_t$ is in the same direction as $\vec{v}$. The resultant $\vec{a}$ points forward-inwards. Angle is $\mathbf{Acute (< 90^\circ)}$.
Slowing down: $\vec{a}_t$ is opposite to $\vec{v}$. Resultant $\vec{a}$ points backward-inwards. Angle is $\mathbf{Obtuse (> 90^\circ)}$.
55.A stone of mass $0.5 \text{ kg}$ is tied to a string of length $1 \text{ m}$ and whirled in a horizontal circle with a uniform speed of $4 \text{ m/s}$. What is the tension in the string? (Assume the string is perfectly horizontal).
Sol: Tension provides necessary centripetal force: $T = F_c = \frac{mv^2}{R}$.
$T = \frac{0.5 \times 4^2}{1} = 0.5 \times 16 = \mathbf{8 \text{ N}}$.
56.Find the radius of curvature of the trajectory of a projectile (launched with speed $u$ at angle $\theta$) at its highest point.
Sol: At highest point, speed $v = u\cos\theta$ (purely horizontal).
Acceleration perpendicular to velocity is full gravity: $a_\perp = g$.
Centripetal accel $a_c = v^2 / R_c \implies g = \frac{(u\cos\theta)^2}{R_c} \implies \mathbf{R_c = \frac{u^2\cos^2\theta}{g}}$.
57.For the same projectile, what is the radius of curvature of the trajectory at the exact point of projection?
Sol: At projection, speed $v = u$.
Gravity $g$ points down. Component of $g$ perpendicular to velocity vector (which is at $\theta$) is $a_\perp = g\cos\theta$.
$a_c = v^2 / R_c \implies g\cos\theta = \frac{u^2}{R_c} \implies \mathbf{R_c = \frac{u^2}{g\cos\theta}}$.
58.A car is travelling at $36 \text{ km/h}$ on a circular road of radius $50 \text{ m}$. It is increasing its speed at the rate of $2 \text{ m/s}^2$. What is its net acceleration?
Sol: $v = 36\text{ km/h} = 10\text{ m/s}$. Tangential acceleration $a_t = 2\text{ m/s}^2$.
Centripetal accel $a_c = \frac{v^2}{R} = \frac{10^2}{50} = \frac{100}{50} = 2\text{ m/s}^2$.
Net acceleration $a = \sqrt{a_t^2 + a_c^2} = \sqrt{2^2 + 2^2} = \mathbf{2\sqrt{2} \text{ m/s}^2}$.
59.Show that for a particle in uniform circular motion, the acceleration vector is always directed towards the center of the circle.
Sol: Position $\vec{r} = R\cos(\omega t)\hat{i} + R\sin(\omega t)\hat{j}$.
Velocity $\vec{v} = \frac{d\vec{r}}{dt} = -R\omega\sin(\omega t)\hat{i} + R\omega\cos(\omega t)\hat{j}$.
Acceleration $\vec{a} = \frac{d\vec{v}}{dt} = -R\omega^2\cos(\omega t)\hat{i} - R\omega^2\sin(\omega t)\hat{j} = -\omega^2(R\cos(\omega t)\hat{i} + R\sin(\omega t)\hat{j})$.
Thus, $\vec{a} = -\omega^2\vec{r}$. The negative sign indicates $\vec{a}$ is anti-parallel to $\vec{r}$ (points strictly towards origin/center).
60.A particle is moving such that its position vector varies with time as $\vec{r} = a\cos(\omega t)\hat{i} + b\sin(\omega t)\hat{j}$. What is the shape of its trajectory? Find the equation of the trajectory.
Sol: $x = a\cos(\omega t) \implies \cos(\omega t) = x/a$.
$y = b\sin(\omega t) \implies \sin(\omega t) = y/b$.
Use $\cos^2(\omega t) + \sin^2(\omega t) = 1 \implies \mathbf{\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1}$.
This represents the standard equation of an Ellipse.
61.Two projectiles are thrown simultaneously in the same vertical plane from the same point. The first is thrown with velocity $v_1$ at angle $\theta_1$ and the second with $v_2$ at $\theta_2$. What is the path of the first particle as seen by the second particle?
Sol: Since both are freely falling, they have same acceleration $\vec{a} = -g\hat{j}$.
Relative acceleration $\vec{a}_{rel} = \vec{a}_1 - \vec{a}_2 = 0$.
Since $\vec{a}_{rel} = 0$, relative velocity $\vec{v}_{rel}$ is constant. An object moving with constant velocity relative to an observer travels in a Straight Line.
62.A hunter aims a gun directly at a monkey sitting on a tree branch. The instant the gun is fired, the monkey drops from the branch. Will the bullet hit the monkey? Prove your answer using relative kinematics.
Sol: Let bullet be B and monkey be M. Initial relative velocity $\vec{v}_{BM,0}$ points directly from gun to monkey.
Accelerations: $\vec{a}_B = \vec{g}$ and $\vec{a}_M = \vec{g}$.
Relative acceleration $\vec{a}_{rel} = \vec{g} - \vec{g} = 0$.
Since relative acceleration is 0, the bullet travels in a straight line relative to the monkey along the initial line of sight. Yes, it will always hit.
63.An elevator is moving upward with a constant acceleration $a_0$. A person inside the elevator throws a ball horizontally with a speed $u$ relative to the elevator. Determine the horizontal range of the ball relative to the elevator floor, given the throw height is $h$.
Sol: In elevator frame, pseudo acceleration is $a_0$ downwards. Effective gravity $g_{eff} = g + a_0$.
Time to fall height $h$: $t = \sqrt{\frac{2h}{g_{eff}}} = \sqrt{\frac{2h}{g+a_0}}$.
Horizontal range $R = u_x t = \mathbf{u\sqrt{\frac{2h}{g+a_0}}}$.
64.A particle is projected from the origin in the x-y plane. Its velocity at time $t$ is given by $\vec{v} = (a)\hat{i} + (b - ct)\hat{j}$. Find the maximum y-coordinate reached by the particle.
Sol: $v_y = b - ct$. Integrate to find y: $y = \int (b - ct)dt = bt - \frac{ct^2}{2}$.
Max height occurs when $v_y = 0 \implies b - ct = 0 \implies t = b/c$.
Substitute $t$: $y_{max} = b(b/c) - \frac{c(b/c)^2}{2} = \frac{b^2}{c} - \frac{b^2}{2c} = \mathbf{\frac{b^2}{2c}}$.
65.A cannon ball is fired to hit a target at a horizontal distance $R$. However, it falls short by $x$ when fired at angle $\alpha$. It overshoots by $y$ when fired at angle $\beta$. Find the correct angle of projection to hit the target exactly. (Assume same initial velocity).
Sol: Range formulas: $R - x = \frac{u^2}{g}\sin 2\alpha$ and $R + y = \frac{u^2}{g}\sin 2\beta$.
Multiply first by $y$, second by $x$, and add to eliminate $R$:
$y(R-x) + x(R+y) = yR + xR = R(x+y)$.
Also $y(\frac{u^2}{g}\sin 2\alpha) + x(\frac{u^2}{g}\sin 2\beta) = \frac{u^2}{g}(y\sin 2\alpha + x\sin 2\beta)$.
Thus, true $R = \frac{u^2}{g} \left( \frac{y\sin 2\alpha + x\sin 2\beta}{x+y} \right)$. Since true $R = \frac{u^2}{g}\sin 2\theta$, we get $\mathbf{\sin 2\theta = \frac{y\sin 2\alpha + x\sin 2\beta}{x+y}}$.
66.Two identical balls A and B are projected from a cliff of height $H$. Ball A is projected horizontally with speed $u$. Ball B is dropped from rest. If both hit the ground at times $t_A$ and $t_B$ respectively, compare $t_A$ and $t_B$.
Sol: Both balls have initial vertical velocity $u_y = 0$.
Vertical motion equation: $H = u_y t + \frac{1}{2}gt^2 = \frac{1}{2}gt^2$.
Thus $t = \sqrt{2H/g}$ for both. Therefore, $\mathbf{t_A = t_B}$.
67.A boy on a flat car moving with constant velocity $10 \text{ m/s}$ throws a ball directly upward with speed $5 \text{ m/s}$ relative to the car. What is the radius of curvature of the ball's path as seen by a stationary observer on the ground, at the highest point of its trajectory?
Sol: For ground observer, velocity at highest point is purely horizontal: $v = v_x = 10\text{ m/s}$.
Perpendicular acceleration is gravity $a_\perp = g = 10\text{ m/s}^2$.
$R_c = \frac{v^2}{a_\perp} = \frac{10^2}{10} = \frac{100}{10} = \mathbf{10 \text{ m}}$.
68.A particle moves in a plane with constant acceleration. The initial velocity is $\vec{v}_0$ and the acceleration is $\vec{a}$. If the particle's speed is minimum at $t = t_0$, find the expression for $t_0$.
Sol: Velocity $\vec{v}(t) = \vec{v}_0 + \vec{a}t$.
Speed squared $v^2 = (\vec{v}_0 + \vec{a}t) \cdot (\vec{v}_0 + \vec{a}t) = v_0^2 + a^2t^2 + 2(\vec{v}_0 \cdot \vec{a})t$.
To minimize speed, differentiate wrt $t$ and set to 0:
$\frac{d(v^2)}{dt} = 2a^2t + 2(\vec{v}_0 \cdot \vec{a}) = 0 \implies \mathbf{t_0 = -\frac{\vec{v}_0 \cdot \vec{a}}{a^2}}$.
69.Find the minimum initial velocity required for a projectile to cross a vertical wall of height $h$ situated at a horizontal distance $x$ from the point of projection.
Sol: Trajectory: $y = x\tan\theta - \frac{gx^2}{2u^2}(1+\tan^2\theta)$. Set $y=h$.
Rearrange as a quadratic in $\tan\theta$: $\left(\frac{gx^2}{2u^2}\right)\tan^2\theta - x\tan\theta + \left(h + \frac{gx^2}{2u^2}\right) = 0$.
For real angle $\theta$, discriminant $D \ge 0 \implies x^2 - 4\left(\frac{gx^2}{2u^2}\right)\left(h + \frac{gx^2}{2u^2}\right) \ge 0$.
Divide by $x^2$: $1 - \frac{2g}{u^2}\left(h + \frac{gx^2}{2u^2}\right) \ge 0 \implies u^4 - 2ghu^2 - g^2x^2 \ge 0$.
Solve quadratic inequality for $u^2$: $u^2 \ge \frac{2gh + \sqrt{4g^2h^2 + 4g^2x^2}}{2} = g(h + \sqrt{h^2+x^2})$.
Thus, $\mathbf{u_{min} = \sqrt{g(h + \sqrt{x^2+h^2})}}$.
70.A particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a_c$ varies with time $t$ as $a_c = k^2 r t^2$, where $k$ is a constant. Find the power delivered to the particle by the forces acting on it.
Sol: $a_c = \frac{v^2}{r} = k^2 r t^2 \implies v^2 = k^2 r^2 t^2 \implies v = k r t$.
Tangential acceleration $a_t = \frac{dv}{dt} = k r$.
Tangential force $F_t = m a_t = m k r$. (Centripetal force does no work).
Power $P = F_t \cdot v = (m k r) (k r t) = \mathbf{m k^2 r^2 t}$.
