First slide

Projection Under uniform Acceleration

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

Easy

A

1200 m
B

0.33 km
C

3.33 km
D

33 km

Solution

$R = v \times t = v \times \sqrt{\frac{2 h}{g}} \begin{array}{l} = \frac{500}{3} \times 20 \\ = 3 .3 km \end{array}$

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