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Q.
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_______
The given situation is shown in the following figure.
If time taken by the ball to reach at highest point $A$ is $t$, then
$0=v-g t \Rightarrow t=\frac{V}{g}$
Total time taken to reach the ball from point of projection to reach at point $B$ is given as
$t_{1}=t +t=2 t \Rightarrow t_{1}=\frac{2 v}{g}$
If $t_{2}$ be the time taken by the ball to reach from point $B$ to point $C$, then
$h=v t_{2}+\frac{1}{2} g t_{2}^{2}$
$\Rightarrow g t_{2}{ }^{2}+2 v t_{2}-2 h=0$
$\therefore t_{2}=\frac{-2 v \pm \sqrt{4 v^{2}+8 g h}}{2 g}$
$t_{2}=\frac{-v}{g}+\sqrt{\frac{v^{2}+2 g h}{2}}$
[taking + ve sign because time is positive]
$\therefore $ Total time,
$T =t_{1}+t_{2}=\frac{2 v}{g}-\frac{v}{g}+\sqrt{\frac{v^{2}+2 g h}{2}}$
$=\frac{v}{g}+\frac{v}{g} \sqrt{1+\frac{2 g h}{v^{2}}}=\frac{v}{g}\left[1+\sqrt{1+\frac{2 g h}{v^{2}}}\right]$
