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If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is
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- (u + gt) t
- ut
- 1/2 gt²
- ut - 1/2 gt²
Correct Option: C
Let body takes T sec to reach maximum height.
Then v = u – gT
v = 0, at highest point. T = u/g …(1)
Velocity attained by body in (T – t) sec v = u – g (T – t)
= u – gT + gt = u – g | + gt | |
g |
or v = gt …(2)
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∴ Distance travelled in last t sec of its ascent
s = (gt)t - | gt² = | gt² | ||
2 | 2 |