Q. Two bodies are thrown simultaneously from a tower with same initial velocity $v_{0}$: one vertically upwards, the other vertically downwards. The distance between the two bodies after time $t$ is
A) $2v_{0}t + \frac{1}{2}gt^{2}$
B) $2v_{0}t$
C) $v_{0}t + \frac{1}{2}gt^{2}$
D) $v_{0}t$
Explanation
(b) For vertically upward motion, $h_{1} = v_{0}t - \frac{1}{2}gt^{2}$ and for vertically downward motion, $h_{2} = v_{0}t + \frac{1}{2}g\ t^{2}$
$\therefore Total\ distance\ covered\ in\ t\sec{h = h_{1} + h_{2} = 2v_{0}t}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.