Question

A bomb weighing $3.7\text{kg}$ is dropped from an aeroplane flying horizontally at $100{\text{ms}}^{-1}$. If the plane is $1\text{km}$ high, what is the horizontal distance that the bomb travels before reaching the ground?
(Take $g=10m/s^2$)

• A. $37000\text{m}$
• B. $3700\text{m}$
• C. $37\times {10}^9\text{m}$
• D. $1000\sqrt{2}\text{m}$

Answer 1

The correct option is D $1000\sqrt{2}\text{m}$

Height of the plane from the ground $=1000\text{m}$
Using second equation of motion,
$h=u_{y}t+\frac{1}{2}g{t}^2$
$\Rightarrow t=\sqrt{\frac{2h}{g}}$
$\Rightarrow t=\sqrt{\frac{2\times 1000}{10}}$
Horizontal distance that the bomb travels before reaching the ground
$x=v_x\times t$
$\Rightarrow x=100\times \sqrt{\frac{2\times 1000}{10}}=1000\sqrt{2}\text{m}$

Alternate solution:

Horizontal distance covered by a body falling from from height $h$ with initial horizontal velocity $u$
$x=u\sqrt{\frac{2h}{g}}=100\times \sqrt{\frac{2\times 1000}{10}}=1000\sqrt{2}\text{m}$