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Question

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

(Take g=10 m/s2)

A
5t2
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B
t2
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C
3t2
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D
8t2
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Solution

The correct option is A 5t2
Let, the maximum height of flight is H and the time of flight is T=u/g
So, H=uT12gT2
Now the distance covered by the ball in time (Tt) sec is
h=u(Tt)12g(Tt)2
So, the distance covered in last t sec is
d=Hh=uT12gT2uT+ut+12gT2gTt+12gt2=utgTt+12gt2=utg(u/g)t+12gt2=12gt2

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