Question

A ball is thrown straight upward with a speed $v$ from a point $h$ meters above the ground. The time taken for the ball to strike the ground is

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

Answer 1

The correct option is A $\frac{v}{g}[1+\sqrt{1+\frac{2hg}{v^2}}]$

$s=vt-\frac{1}{2}g{t}^2$
Now, at $C$ (maximum height point),
$h^{‘}=\frac{v^2}{2g}$
So, for journey from $C$ to $B$,
time $=\sqrt{\frac{2(h+h^{‘})}{g}}=\sqrt{\frac{2(h+\frac{v^2}{2g})}{g}}=t_2$
$\Rightarrow$ total time $=t_1+t_2=\frac{v}{g}(1+\sqrt{1+\frac{2hg}{v^2}})$