Question

A body thrown vertically up to reach its maximum height in $t$ second. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is

• A. $\sqrt{2} \, t $
• B. $ 1 + \frac{1}{\sqrt{2}} t $
• C. $\frac{3t}{2} $
• D. $\frac{t}{\sqrt{2}} $

Answer 1

Answer: (B)

The ball is thrown vertically upwards then according to equation of motion
$(0)^2 - u^2 = -2gh$ ...(i)
and $0 = u - gt$ ....(ii)
From Eqs. (i) and (ii),
$h = \frac{gt^2}{2} $
When the ball is falling downwards after reaching the maximum height
$s = ut' + \frac{1}{2} g(t')^2$
$\frac{h}{2} = (0) t' + \frac{1}{2} g (t')^2$
$\Rightarrow \, \, t' = \frac{\sqrt{h}}{g}$
$ t ' = \frac{t}{\sqrt{2}}$
Hence, the total time from the time of projection to reach a point at half of its maximum height while returning = $t + t'$
$ = t + \frac{t}{\sqrt{2}}$