If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is |
$\frac{1}{2}gt^2$ $ut-\frac{1}{2}gt^2$ $(u-gt)t$ $ut$ |
$\frac{1}{2}gt^2$ |
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 $S_{n-t}=u(n-t)-\frac{1}{2}g(n-t)^2$ $∴ s =S_n-S_{n-t}=\frac{1}{2}gt^2$ |