First slide

Projection Under uniform Acceleration

Question

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

Moderate

A

$2 v \sqrt{(h / g)}$
B

$V \sqrt{(h / g)}$
C

$V \sqrt{(2 h / g)}$
D

$V \sqrt{(h / 2 g)}$

Solution

The time taken by the bomb to hit the ground is given by
$\begin{array}{l} h = 0 \times t + \frac{1}{2} {gt}^{2} or (2 h / g) = t^{2} \\ ∴ t = \sqrt{(2 h / g)} \end{array}$
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
$= vt = v \sqrt{(2 h / g)}$

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