Question

An aeroplane is flying horizontally with a velocity of 600 km/h at a height of 1960 m. When it is vertically at a point A on the ground, a bomb is released from it. The bomb strikes the ground at point B. The distance AB is

• A. 1200 m
• B. 0.33 km
• C. 3.33 km
• D. 33 km

Answer 1

The correct option is C. 3.33 km.

The velocity with which the bomb is released is 600 km/h (horizontal).

Vertical velocity of the bomb is zero. Therefore using the second equation of motion we get,

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

$⟹h=0+\frac{1}{2}g{t}^2$

$⟹t=\sqrt{\frac{2h}{g}}$

Horizontal displacement of the bomb
AB = Horizontal velocity $\times$ time available
$AB=u\times \sqrt{\frac{2h}{g}}=600\times \frac{5}{18}\times \sqrt{\frac{2\times 1960}{9.8}}=3.33km$.