A bomb is dropped from an aeroplane when it is at a height h directly above a target. If the aeroplane is moving horizontally at a speed v, the distance by
which the bomb will miss the target is given by

The time taken by the bomb to hit the ground is given by
$\begin{array}{l}\mathrm{h}=0\times \mathrm{t}+\frac{1}{2}{\mathrm{gt}}^2 \mathrm{or} (2\mathrm{h}/\mathrm{g})={\mathrm{t}}^2\\ \therefore \hspace{0.25em}\mathrm{t}=\sqrt{(2\mathrm{h}/\mathrm{g})}\end{array}$
Horizontal speed of aeroplane = Horizontal speed of bomb (= v)
The distance by which the bomb will miss the target is given by
= horizontal velocity , time
$=\mathrm{vt}=\mathrm{v}\sqrt{(2\mathrm{h}/\mathrm{g})}$