Question

A ball is projected upwards from a height $h$ above the surface of the earth with velocity $v$. The time at which the ball strikes the around is_______

• A. $\frac{v}{g}\left[1+\frac{\sqrt{2gh}}{v^2}\right]$
• B. $\frac{v}{g}\left[1-\sqrt{1+\frac{2h}{g}}\right]$
• C. $\frac{v}{g}\left[1+\sqrt{1+\frac{2gh}{v^2}}\right]$
• D. $\frac{v}{g}\left[1+\sqrt{v^2+\frac{2g}{v^2}}\right]$

Answer 1

Answer: (C)

The given situation is shown in the following figure.

If time taken by the ball to reach at highest point $A$ is $t$, then
$0=v-gt\Rightarrow t=\frac{V}{g}$
Total time taken to reach the ball from point of projection to reach at point $B$ is given as
$t_1=t+t=2t\Rightarrow t_1=\frac{2v}{g}$
If $t_2$ be the time taken by the ball to reach from point $B$ to point $C$, then
$h=v{t}_2+\frac{1}{2}g{t}_2^2$
$\Rightarrow g{t}_2{}^2+2v{t}_2-2h=0$
$\therefore t_2=\frac{-2v\pm \sqrt{4v^2+8gh}}{2g}$
$t_2=\frac{-v}{g}+\sqrt{\frac{v^2+2gh}{2}}$
[taking + ve sign because time is positive]
$\therefore$ Total time,
$T=t_1+t_2=\frac{2v}{g}-\frac{v}{g}+\sqrt{\frac{v^2+2gh}{2}}$
$=\frac{v}{g}+\frac{v}{g}\sqrt{1+\frac{2gh}{v^2}}=\frac{v}{g}\left[1+\sqrt{1+\frac{2gh}{v^2}}\right]$