Question

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

• A.
  $\frac{1}{2}g{t}^2$
• B.
  $ut-\frac{1}{2}g{t}^2$
• C. $(u-gt)t$
• D. $utd$

Answer 1

The correct option is A

$\frac{1}{2}g{t}^2$

The distance covered bt the ball during the last t seconds of its upward motion =Distance covered by it in first t seconds of its downward motion.

From $h=ut+\frac{1}{2}g{t}^2$
$h=\frac{1}{2}g{t}^2$[As u = 0 for it downward motion ]