Q. An aeroplane is flying horizontally with a velocity of $216\ kmh^{- 1}$ 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 (ignoring air resistance)

A) 1200 m
B) 0.33 km
C) 3.33 km
D) 33 km
Explanation

(a) Horizontal velocity of aeroplane,
$u = \frac{216 \times 1000}{60 \times 100} = 60\ ms^{- 1}$
Time of flight, $T = \sqrt{\frac{2s}{g}} = \sqrt{\frac{2 \times 1960}{9.8}} = 20s$
Horizontal range, $AB = uT$
$= 60 \times 20 = 1200m$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.