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