Question

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

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

Answer 1

Answer: (A)

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

Answer 2

If ball is thrown with velocity u, then time of flight $= \frac{u}{g}$

velocity after $\left( \frac{u}{g} - t \right)\sec:$ $v = u - g\left( \frac{u}{g} - t \right)$ = gt.

So, distance in last 't' sec : $0^{2} = (gt)^{2} - 2(g)h.$

⇒ $h = \frac{1}{2}gt^{2}.$