question_answer

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

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                 \[CB=v't-\frac{1}{2}\,g{{t}^{2}}\]                 \[=(gt)\,t-\frac{1}{2}\,g{{t}^{2}}\]                 \[=g{{t}^{2}}-\frac{1}{2}g{{t}^{2}}[From\,Eq.\,(ii)]\]                 \[=\frac{1}{2}g{{t}^{2}}\]