Given, speed of packets = 125 m/s

Height of the hill = 500 m.

To cross the hill, the vertical component of the velocity should be sufficient to cross such height.

$u_y \geq \sqrt{2gh}$

$\geq \sqrt{2 \times 10 \times 500}$

$\geq 100 \text{m/s}$

But

$u^2 = u_x^2 + u_y^2$

$\therefore$ Horizontal component of initial velocity,

$u_x = \sqrt{u^2 - u_y^2} = \sqrt{(125)^2 - (100)^2} = 75 \text{ m/s}$

Time taken to reach the top of the hill,

$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 500}{10}} = 10 \text{ s}$

Time taken to reach the ground from the top of the hill $t' = t = 10 \text{ s}$. Horizontal distance travelled in 10 s

$x = u_x \times t = 75 \times 10 = 750 \text{ m}$

$\therefore$ Distance through which canon has to be moved = 800 – 750 = 50 m

Speed with which canon can move = 2 m/s

$\therefore$ Time taken by canon = $\frac{50}{2}$ $\Rightarrow$ $t'' = 25 \text{ s}$

$\therefore$ Total time taken by a packet to reach on the ground = $t'' + t + t' = 25 + 10 + 10 = 45$ s

This problem can be approached in two different ways

(i) Refer to the diagram, target $T$ is at horizontal distance $x = R + \Delta x$ and between point of projection $y = -h$.

(ii) From point $P$ in the diagram projection at speed $v_0$ at an angle $\theta$ below horizontal with height $h$ and horizontal range $\Delta xA$)

Applying method (i)

Maximum horizontal range

$$R = \frac{{v_0^2}}{g}, \text{ for } \theta = 45^\circ\quad \text{...(i)}$$

Let the gun be raised through a height h from the ground so that it can hit the target. Let vertically downward direction be taken as positive.

Horizontal component of initial velocity = v₀ cosθ

Vertical component of initial velocity = −v₀ sinθ

Taking motion in vertical direction,

$ h = (-v_0 \sin\theta) \, t + \frac{1}{2} \, g \, t^2 $ ... (ii)

Taking motion in horizontal direction

$(R + Δx) = v_0 \cos\theta \times t $

$\Rightarrow$ $ t = \frac{(R + Δx)}{v_0 \cos\theta} $ ... (iii)

Substituting value of t in Eq. (ii), we get

$ \begin{align*} h &= \left(-v_0 \sin\theta \right) \times \left( \frac{R + Δx}{v_0 \cos\theta} \right) + \frac{1}{2} \, g \, \left( \frac{R + Δx}{v_0 \cos\theta} \right)^2 \\ h &= - (R + Δx) \tan\theta + \frac{1}{2} \, g \, \frac{(R + Δx)^2}{v_0^2 \cos^2\theta} \end{align*} $

$\begin{align*} &\text{As angle of projection is } \theta = 45^\circ, \text{ therefore} \\\\ & h = -(R + \Delta x) + \tan 45^\circ + \frac{1}{2} \frac{g\ (R + \Delta x)^2}{v_0^2 \cos^2 45^\circ} \\\\ & h = -(R + \Delta x) \times 1 + \frac{1}{2} \frac{g \ (R + \Delta x)^2}{v_0^2 (1/2)} \quad \left(\because \tan 45^\circ = 1 \text{ and } \cos 45^\circ = \frac{1}{\sqrt{2}}\right) \\\\ & h = -(R + \Delta x) + \frac{(R + \Delta x)^2}{R} \quad \text{[Using Eq. (i), } R = \frac{v_0^2}{g}] \\\\ & = -(R + \Delta x) + \frac{1}{R}(R^2 + \Delta x^2 + 2R\Delta x) \\\\ & = -R - \Delta x + \left( R + \frac{\Delta x^2}{R} + 2\Delta x \right) \\\\ & = \Delta x + \frac{\Delta x^2}{R} \\\\ & h = \Delta x \left( 1 + \frac{\Delta x}{R} \right) \\\\ & \text{Hence proved.} \end{align*}$

Note We should not confuse with the positive direction of motion. May be vertically upward direction or vertically downward direction is taken as positive according to convenience.

Consider the adjacent diagram.

Mutually perpendicular x and y-axes are shown in the diagram.

Particle is projected from the point O.

Let time taken in reaching from point O to point P is T.

(b) Considering motion along vertical upward direction perpendicular to OX.

For the journey O to P.

$ \begin{align*} & y = 0, \ u_y = v_0 \sin \beta, \ a_y = -g \cos \alpha, \ t = T \\ & \text{Applying equation,} \\ & y = u_y t + \frac{1}{2} a_y t^2 \\ & \implies 0 = v_0 \sin \beta T + \frac{1}{2}(-g \cos \alpha) T^2 \end{align*} $

$\Rightarrow \ T \left[ v_0 \sin \beta - \frac{g \cos \alpha}{2} T \right] = 0 $

$\Rightarrow \quad T = 0, \ T = \frac{2\ v_0 \sin \beta}{g \cos \alpha} $

As $ T = 0 $, corresponding to point $ O $

Hence, $T = \text{Time of flight} = \frac{2\ v_0 \sin \beta}{g \cos \alpha} $

(a)$\text{ Considering motion along } OX.$

$ x = L, u_x = v_0 \cos \beta, a_x = -g \sin \alpha $

$ t = T = \frac{2\ v_0 \sin \beta}{g \cos \alpha} $

$ x = u_x t + \frac{1}{2} a_x t^2 $

$\begin{align*} \implies & \quad L = v_0 \cos \beta \ T + \frac{1}{2} (-g \sin \alpha) \ T^2 \\\\ \implies & \quad L = v_0 \cos \beta \ T - \frac{1}{2} g \sin \alpha \ T^2 \\\\ & \quad = T \left[v_0 \cos \beta - \frac{1}{2} g \sin \alpha \ T \right] \\\\ & \quad = T \left[ v_0 \cos \beta - \frac{1}{2} g \sin \alpha \times \frac{2 v_0 \sin \beta}{g \cos \alpha} \right] \\\\ & \quad = \frac{2 \ v_0 \sin \beta}{g \cos \alpha} \left[ v_0 \cos \beta - \frac{v_0 \sin \alpha \sin \beta}{\cos \alpha} \right] \\\\ & \quad = \frac{2 \ v_0^2 \sin \beta}{g \cos^2 \alpha} [\cos \beta \cdot \cos \alpha - \sin \alpha \cdot \sin \beta] \\\\ \implies & \quad L = \frac{2 \ v_0^2 \sin \beta}{g \cos^2 \alpha} \cos (\alpha + \beta) \end{align*}$

$(c)$ For range $(L)$ to be maximum, $ \sin \beta \cdot \cos (\alpha + \beta) $ should be maximum.

Let,

$ \begin{align*} Z &= \sin \beta \cdot \cos (\alpha + \beta) \\ &= \sin \beta \left[ \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta \right] \\ &= \frac{1}{2} \left[ \cos \alpha \cdot \sin 2\beta - 2 \sin \alpha \cdot \sin^2 \beta \right] \\ &= \frac{1}{2} \left[ \sin 2 \beta \cdot \cos \alpha - \sin \alpha \left(1 - \cos 2 \beta\right) \right] \\ \implies z &= \frac{1}{2} \left[ \sin 2 \beta \cdot \cos \alpha - \sin \alpha + \sin \alpha \cdot \cos 2 \beta \right] \\ &= \frac{1}{2} \left[ \sin 2 \beta \cdot \cos \alpha + \cos 2 \beta \cdot \sin \alpha - \sin \alpha \right] \\ &= \frac{1}{2} \left[ \sin (2 \beta + \alpha) - \sin \alpha \right] \end{align*} $

For $z$ to be maximum,

$ \sin (2 = \beta + \alpha) = \text{maximum} = 1 $

$ \implies 2 \beta + \alpha = \frac{\pi}{2} \text{ or, } \beta = \frac{\pi}{4} - \frac{\alpha}{2} $

Considering x and y-axes as shown in the diagram. For the motion of the projectile from O to A.

$ y = 0, u_y = v_0 \cos \theta \\ a_y = -g \cos \theta, t = T $

Applying equation of kinematics,

$ y = u_y t + \frac{1}{2} a_y t^2 $

$ ⇒ \quad 0 = v_0 \cos \theta \; T + \frac{1}{2}(-g \cos \theta) T^2 $

$ ⇒ \quad T \left[ v_0 \cos \theta - \frac{g \cos \theta \; T}{2} \right] = 0 $

$ T = \frac{2 v_0 \cos \theta}{g \cos \theta} $

As $T = 0$, corresponds to point $O$

Hence,

$ T = \frac{2v_0}{g} $

Now considering motion along $OX$.

$ x = L, \ u_x = v_0 \sin \theta, \ a_x = g \sin \theta, \ t = T = \frac{2v_0}{g} $

Applying equation of kinematics,

$ x = u_x t + \frac{1}{2} a_x t^2 $

$ \Rightarrow L = v_0 \sin \theta \ t + \frac{1}{2} g \sin \theta \ t^2 = (v_0 \sin \theta) (T) + \frac{1}{2} g \sin \theta \ T^2 $

$ \qquad = (v_0 \sin \theta) \left( \frac{2v_0}{g} \right) + \frac{1}{2} g \sin \theta \times \left( \frac{2v_0}{g} \right)^2 $

$ \qquad = \frac{2v_0^2}{g} \sin \theta + \frac{1}{2} g \sin \theta \times \frac{4v_0^2}{g^2} = \frac{2v_0^2}{g} [\sin \theta + \sin \theta] $

$ \Rightarrow L = \frac{4v_0^2}{g} \sin \theta $

Assume north to be $ \hat{i} $ direction and vertically downward to be $-\hat{j} $.

Let the rain velocity $ \mathbf{v}_r $ be $ a \hat{i} + b \hat{j} $.

$\mathbf{v}_r = a \hat{i} + b \hat{j} $

Case I Given velocity of girl = $ \mathbf{v}_g = (5 \text{ m/s})\hat{i} $

Let $ \mathbf{v}_{rg} $ = Velocity of rain w.r.t girl

$ \mathbf{v}_{rg} = \mathbf{v}_r - \mathbf{v}_g = (a\hat{i} + b\hat{j}) - 5\hat{i} $

$ = (a - 5) \hat{i} + b \hat{j} $

According to question rain, appears to fall vertically downward. Hence,

$ a - 5 = 0 \Rightarrow a = 5 $

Case II Given velocity of the girl, $ \mathbf{v_g} = (10 \text{ m/s}) \hat{i} $

$ \mathbf{v_{rg}} = \mathbf{v_r} - \mathbf{v_g} $

$ = (a\hat{i} + b\hat{j}) - 10\hat{i} = (a - 10)\hat{i} + b\hat{j} $

According to question rain appears to fall at 45° to the vertical hence $ \tan 45° = \frac{b}{a - 10} = 1 $

$ \Rightarrow b = a - 10 = 5 - 10 = -5 $

Hence, velocity of rain $ = a\hat{i} + b\hat{j} $

$ \Rightarrow \mathbf{v_r} = 5\hat{i} - 5\hat{j} $

Speed of rain = $|\mathbf{v_r}| = \sqrt{(5)^2 + (-5)^2} = \sqrt{50} = 5\sqrt{2} \text{ m/s} $