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Q.
A ball is thrown straight upward with a velocity $v_{0}$ from a height $h$ above the ground. The time taken for the ball to strike the ground is
Motion in a Straight Line
Solution:
Apply $s=u t+\frac{1}{2} a t^{2}$
$\Rightarrow -h=v_{0} t-\frac{1}{2} g t^{2}$
$\Rightarrow g t^{2}-2 v_{0} t-2 h=0$
$\Rightarrow t=\frac{2 v_{0} \pm \sqrt{4 v_{0}^{2}+8 g h}}{2 g}$
As $t$ is not negative, so ignoring negative sign, we get
$t=\frac{2 v_{0}+2 \sqrt{v_{0}^{2}+2 g h}}{2 g}$
$=\frac{v_{0}}{g}\left[1+\sqrt{1+\frac{2 g h}{v_{0}^{2}}}\right]$