Q. If a ball is thrown vertically upwards with speed $u$, the distance covered during the last $t$ seconds of its ascent is
A) $\frac{1}{2}gt^{2}$
B) $ut - \frac{1}{2}gt^{2}$
C) $(u - gt)t$
D) $ut$
Explanation
(a) The distance covered by the ball during the last $t\ seconds$ of its upward motion $=$ Distance covered by it in first $t\ $seconds of its downward motion
From $h = = ut + \frac{1}{2}gt^{2}$
$h = \frac{1}{2}g\ t^{2}$ [As $u = 0$ for it downward motion]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.