If a ball is thrown vertically upwards with speed `u`, the distance covered during the last `t` second of its ascent is

Explanation

Motion from `A` to `B`: velocity at the highest point`=0`
`0=v-g t_(0)impliesv=g t_(0)`
`h=v t_(0)-1/2 g t_(0)^(2)=g t_(0).t_(0)-1/2g t_(0)^(2)=1/2g t_(0)^(2)`
OR
Displacement in the first `t_(0)` second of the downward journey
`h=0+1/2g t_(0)^(2)=1/2 g t_(0)^(2)`
OR
Recall: If the particle is moving in a straight line under constant acceleration//retardation and velocity at one end is zero (initial or final), displacement in time `t, s =1/2at^(2)`, hence, `h=1/2 g t_(0)^(2)`.