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 ground is

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Aicharya · Accepted Answer

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 $\begin{aligned} h & =v t_2+\frac{1}{2} g t_2^2 \Rightarrow g t_2^2+2 v t_2-2 h=0 \\ t_2 & =\frac{-2 v \pm \sqrt{4 v^2+8 g h}}{2 g} \\ t_2 & =\frac{-v}{g}+\sqrt{\frac{v^2+2 g h}{2}} \end{aligned}$ [taking +ve sign because time is positive] $\therefore$ Total time, $T=t_1+t_2=\frac{2 v}{g}-\frac{v}{g}+\sqrt{\frac{v^2+2 g h}{2}}$ $=\frac{v}{g}+\frac{v}{g} \sqrt{1+\frac{2 g h}{v^2}}=\frac{v}{g}\left[1+\sqrt{1+\frac{2 g h}{v^2}}\right]$

A ball is projected upwards from a height \(h\) above the surface of the earth with velocity \(v\). The time…

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