Q. 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
A) $1200\ m$
B) $0.33\ km$
C) $3.33\ km$
D) $33\ km$
Explanation
(c) Horizontal displacement of the bomb
$AB =$Horizontal velocity$\ \times \ $time available
$AB = u \times \sqrt{\frac{2h}{g}} = 600 \times \frac{5}{18} \times \sqrt{\frac{2 \times 1960}{9.8}} = 3.33\ km$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.