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 ground is
A) $\frac{v}{g} + \frac{2hg}{\sqrt{2}}$
B) $\frac{v}{g}\left\lbrack 1 - \sqrt{1 + \frac{2h}{g}} \right\rbrack$
C) $\frac{v}{g}\left\lbrack 1 + \sqrt{1 + \frac{2gh}{v^{2}}} \right\rbrack$
D) $\frac{v}{g}\left\lbrack 1 + \sqrt{v^{2} + \frac{2h}{g}} \right\rbrack$
Explanation
(c) Since direction of $v$ is opposite to the direction of $g$ and $h$ so from equation of motion
$h = - vt + \frac{1}{2}gt^{2} \Rightarrow gt^{2} - 2vt - 2h = 0$
$\Rightarrow t = \frac{2v \pm \sqrt{4v^{2} + 8gh}}{2g} \Rightarrow t = \frac{v}{g}\left\lbrack 1 + \sqrt{1 + \frac{2gh}{v^{2}}} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.