Q. An aeroplane is flying horizontally with a velocity of $600\ {kmh}^{- 1}$ 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
A) 1200 m
B) 0.33 km
C) 333.3 km
D) 3.33 km
Explanation
(d) $h = \frac{1}{2}gt^{2}\ \Rightarrow t = \sqrt{\frac{2 \times 1960}{9.8}} = 20\ s$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = AB = ut = 600 \times \frac{20}{60 \times 60}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3.33\ km$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.