If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is
(a) (u + gt)t
(b) ut
(c) 1/2 gt^2
(d) ut-1/2 gt^2

(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 + zt = u-g u/g + gt
or v = gt                 …………. (2)
 Distance travelled in last t sec of its ascent
S = (gt)t -1/2 gt2 = 1/2  gt2

See, distance covered in the last t seconds of ascent is equal to the distance covered in the first t seconds of descent. So we have to find the distance of first t descent. s=ut+1/2at^2Here, a=g (it will be taken positive as we are taking it with the gravity)u=0(as at top the velocity will be zero and it will be the initial velocity of descent)So,s=0.t+1/2gt^2And, s=1/2gt^2.

Hello Student

Please see the solution in the attachment.

So option C is correct.

I hope this solution will help you.