question_answer

If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is:                        [AIPMT 2003]

A) \[ut-\frac{1}{2}g\,{{t}^{2}}\]

B) \[(u+gt)t\]

C) \[ut\]

D) \[\frac{1}{2}g{{t}^{2}}\]

Correct Answer: D

Solution :

Let the ball takes T sec to reach maximum height H.

\[v=u-gT\]

put \[v=0\] (at height H)

\[\therefore \]      \[u=gT\] or \[T=u/g\]                  (i)

Velocity attained by the ball in

\[(T-t)\] sec is,

\[v'=u-g\,(T-t)\]

\[=u-gT+gt\]

\[=u-g\frac{u}{g}+gt\]

\[=u-u+gt\]

\[v=gt\]                                     (ii)

Hence, distance travelled in last t sec of its ascent

\[CB=v't-\frac{1}{2}\,g{{t}^{2}}\]

\[=(gt)\,t-\frac{1}{2}\,g{{t}^{2}}\]

\[=g{{t}^{2}}-\frac{1}{2}g{{t}^{2}}\]                        [From Eq. (ii)]

\[=\frac{1}{2}g{{t}^{2}}\]