Q. A ball is thrown vertically upwards from the top of a tower of height $h$ with velocity $v$. The ball strikes the ground after
A) $\frac{v}{g}\left\lbrack 1 + \sqrt{1 + \frac{2gh}{v^{2}}} \right\rbrack$
B) $\frac{v}{g}\left\lbrack 1 - \sqrt{1 + \frac{2gh}{v^{2}}} \right\rbrack$
C) $\frac{v}{g}\left( 1 + \frac{2gh}{v^{2}} \right)^{1/2}$
D) $\frac{v}{g}\left( 1 - \frac{2gh}{v^{2}} \right)^{1/2}$
Explanation
(a) $h = - vt + \frac{1}{2}gt^{2}$ or $ft^{2} - 2vt - h$=0
$t = \frac{- ( - 2v) \pm \sqrt{4v^{2} + 4gh}}{2g} = \frac{2v \pm 2\sqrt{v^{2} + gh}}{2g}$
$= \frac{v}{g} \pm \frac{\left\lbrack v^{2} + 2gh \right\rbrack^{1/2}}{g} = \frac{v}{g}\left\lbrack 1 \pm \sqrt{1 + \frac{2gh}{v^{2}}} \right\rbrack$
Now, retain only the positive sign.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.