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 \[CB=v't-\frac{1}{2}\,g{{t}^{2}}\] \[=(gt)\,t-\frac{1}{2}\,g{{t}^{2}}\] \[=g{{t}^{2}}-\frac{1}{2}g{{t}^{2}}[From\,Eq.\,(ii)]\] \[=\frac{1}{2}g{{t}^{2}}\]You need to login to perform this action.
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