Question

An aeroplane is travelling horizontally at a height of $2000\text{m}$ from the ground. The aeroplane, when at a point $\text{P}$, drops a bomb to hit a stationary target $Q$ on the ground. In order that the bomb hits the target, what angle $\theta$ must the line $\text{PQ}$ make with the vertical ?
$[g=10{\text{m/s}}^2]$

• A. ${30}^{\circ }$
• B. ${37}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is C ${45}^{\circ }$


Aeroplane when drops the bomb from point $\text{P}$ has a horizontal speed $100\text{m/s}$.

So, Bomb will have initial speed of $100\text{m/s}$ horizontally.


Bomb reaches from $\text{P}\to \text{Q}$ travelling vertical height of $2000\text{m}.$

For vertical motion,

$u_y=0;a_y=g$

Now using second equation of motion,

$s_y=u_{y}t+\frac{1}{2}g{t}^2$

$\Rightarrow h=\frac{1}{2}g{t}^2$

$\Rightarrow 2000=\frac{1}{2}\times 10\times t^2$

$\Rightarrow t=20\text{sec}.$

In the same time $t$ it travels horizontally by a distance $R$

So $R=u_{x}t=100\times 20=2000\text{m}$


$\Rightarrow \tan (90-\theta )=\frac{2000}{2000}=1$

$\Rightarrow 90-\theta ={45}^{\circ }\Rightarrow \theta ={45}^{\circ }$

Hence, $\left(c\right)$ is correct choice.