Consider the adjacent diagram in which vectors A and B are corrected by head and tail. Resultant vector C = A + B

(a) from (iv) it is clear that c = a + b

(b) from (iii) c + b = a → a - c = b

(c) from (i) b = a + c → b - a = c

(d) from (ii) -c = a + b → a + b + c = 0

Given $|A| = 2$ and $|B| = 4$:

(a) $A \cdot B = AB \cos \theta = 0$

$\Rightarrow 2 \times 4 \cos \theta = 0$

$\Rightarrow \cos \theta = 0 = \cos 90^\circ$

$\Rightarrow \theta = 90^\circ$

Thus, option (a) matches with option (ii).

(b) $A \cdot B = AB \cos \theta = 8$

$\Rightarrow 2 \times 4 \cos \theta = 8$

$\Rightarrow \cos \theta = 1 = \cos 0^\circ$

$\Rightarrow \theta = 0^\circ$

Thus, option (b) matches with option (i).

(c) $A \cdot B = AB \cos \theta = 4$

$\Rightarrow 2 \times 4 \cos \theta = 4$

$\Rightarrow \cos \theta = \frac{1}{2} = \cos 60^\circ$

$\Rightarrow \theta = 60^\circ$

Thus, option (c) matches with option (iv).

(d) $A \cdot B = AB \cos \theta = -8$

$\Rightarrow 2 \times 4 \cos \theta = -8$

$\Rightarrow \cos \theta = -1 = \cos 180^\circ$

$\Rightarrow \theta = 180^\circ$

Thus, option (d) matches with option (iii).

Given $|A| = 2$ and $|B| = 4$

(a) $|A \times B| = AB \sin \theta = 0$

$ \Rightarrow 2 \times 4 \sin \theta = 0 $

$ \Rightarrow \sin \theta = 0 = \sin 0^\circ $

$ \Rightarrow \theta = 0^\circ $

$ \therefore \text{Option (a) matches with option (iv).} $

(b) $|A \times B| = AB \sin \theta = 8$

$ \Rightarrow 2 \times 4 \sin \theta = 8 $

$ \Rightarrow \sin \theta = 1 = \sin 90^\circ $

$ \Rightarrow \theta = 90^\circ $

$ \therefore \text{Option (b) matches with option (iii).} $

(c) $|A \times B| = AB \sin \theta = 4$

$ \Rightarrow 2 \times 4 \sin \theta = 4 $

$ \Rightarrow \sin \theta = \frac{1}{2} = \sin 30^\circ $

$ \Rightarrow \theta = 30^\circ $

$ \therefore \text{Option (c) matches with option (i).} $

(d) $|A \times B| = AB \sin \theta = 4 \sqrt{2}$

$ \Rightarrow 2 \times 4 \sin \theta = 4 \sqrt{2} $

$ \Rightarrow \sin \theta = \frac{1}{\sqrt{2}} = \sin 45^\circ $

$ \Rightarrow \theta = 45^\circ $

$ \therefore \text{Option (d) matches with option (ii).} $

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.