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

Options

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

Answer

a

Explanation

Upward motion and downward motion are identical. Hence distance covered during the last t second is same as distance covered in first t are after reaching maximum height.
`S=ut+(1)/(2)at^(2)`
`rArr h=0+(1)/(2)gt^(2)rArrh=(1)/(2)gt^(2)`.