If a ball is thrown vertically upwards with speed $u$, the distance covered during the last $t$ seconds of its ascent is
If a ball is thrown vertically upwards with speed $u$, the distance covered during the last $t$ seconds of its ascent is
$\frac{1}{2}g{t^2}$
$ut - \frac{1}{2}g{t^2}$
$(u - gt)t$
$ut$
If a ball is thrown vertically upwards with speed $u$, the distance covered during the last $t$ seconds of its ascent is
The distance covered by 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]
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