Q. A bomb is dropped from an aeroplane flying horizontally with a velocity $469\ ms^{- 1}$ at an altitude of 980 m. The bomb will hit the ground after a time
A) $2\ s$
B) $\sqrt{2}\ s$
C) $5\sqrt{2}\ s$
D) $10\sqrt{2}\ s$
Explanation
(d) $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 980}{9.8}} = 10\sqrt{2}\ s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.