Question

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

Answer 1

Answer:

3) ut - ½gt²

Explanation:

actually motion equation of distance in time t is

ut + ½at²

but, in when object is thrown upward earth pulls object , due to this retardation occurs and sign of retardation is negative .

so we will use negative acceleration during calculations that is,

S = ut - ½gt²