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

$\frac{1}{2}gt^2$

The correct answer is Option (1) → $\frac{1}{2}gt^2$

The time taken by the ball to reach the highest point

$=u –gn ⇒n=\frac{u}{g}$

$S_n=un-\frac{1}{2}gn^2$

and $S_{n-t}=u(n-t)-\frac{1}{2}g(n-t)^2$

$∴ s =S_n-S_{n-t}=\frac{1}{2}gt^2$