First slide

Projection Under uniform Acceleration

Question

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

Moderate

A

1200 m
B

0.33 km
C

333.3 km
D

3.33 km

Solution

Time of flight of bomb = $\sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 1960}{9.8}} = 20 s$
$AB = (600 \times \frac{5}{18}) (20)$
= 3333.3 m = 3.33 km

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