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

Options

A. ut
B. `1/2"gt"^2`
C. `ut-1/2"gt"^2`
D. (u + gt)t

Answer

B

Explanation

Let time of flight be T then T=`u/g`
Let h be the distance covered during last 't' second of its ascent ltbr gt Velocity at point B= `v_B=u-g(T-t)`
`=u-g(u/g-t)="gt"`

`rArr h=v_Bt-1/2"gt"^ rArr h="gt"^2-1/2"gt"^2=1/2"gt"^2`