Question

A bomb is dropped on an enemy post by an aeroplane flying horizontally with a velocity of $60{\text{km/h}}^{-1}$ and at a height of $490\text{m}$. At the time of dropping the bomb, how far should the aeroplane be from the enemy post so that the bomb may directly hit the target?

• A. $\frac{400}{3}\text{m}$
• B. $\frac{500}{3}\text{m}$
• C. $\frac{1700}{3}\text{m}$
• D. $498\text{m}.$

Answer 1

The correct option is B $\frac{500}{3}\text{m}$

Height of the aeroplane from the ground $=490\text{m}$

Using second equation of motion,
$h=u_{y}t+\frac{1}{2}g{t}^2$
$u_y=0$
$\Rightarrow t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 490}{9.8}}\text{s}=\sqrt{100}\text{s}=10\text{s}$

Let $x$ be the distance between the target and the aeroplane when the bomb is dropping from the aeroplane.
$\Rightarrow x=vt=(60\times \frac{5}{18}){\text{ms}}^{-1}\times 10\text{s}=\frac{500}{3}\text{m}$

Alternate Solution:
Horizontal distance covered by a body falling from height $h$ with initial horizontal velocity $u$.
$x=u\sqrt{\frac{2h}{g}}=60\times \frac{5}{18}\times \sqrt{\frac{2\times 490}{9.8}}=60\times \frac{5}{18}\times 10=\frac{500}{3}\text{m}$