Question

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

• A. $u t$
• B. $\frac{1}{2} g t^{2}$
• C. $u t - \frac{1}{2} g t^{2}$
• D. $\left(\right. u + g t \left.\right) t$

Answer 1

Answer: (B)

Let total height $= H$
Total Time of ascent $= T$
So, $H=uT-\frac{1}{2}gT^{2}$

Distance covered by ball in time $\left(\right. T - t \left.\right) \, \text{sec}$ is
$y = u \left(\right. T - t \left.\right) - \frac{1}{2} g \left(\right. T - t \left.\right)^{2}$
So, distance covered by ball in last $t \, sec$
$h = H - y = \left[u T - \frac{1}{2} g T^{2}\right] - \left[u \left(\right. T - t \left.\right) - \frac{1}{2} g \left(\right. T - t \left.\right)^{2}\right]$
By solving and putting $T = \frac{u}{g}$
we will get,
$h = \frac{1}{2} g t^{2}$