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

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
$AB=u\times \sqrt{\frac{2h}{g}}=600\times \frac{5}{18}\times \sqrt{\frac{2\times 1960}{9.8}}=3.33km$.