Q. A ball thrown upward from the top of a tower with speed $v$ reaches the ground in $t_{1}$ second. If this ball is thrown downward from the top of he same tower with speed $v$ it reaches the ground in $t_{2}$ second. In what time the ball shall reach the ground if it is allowed to falls freely under gravity from the top of the tower?
A) $\frac{t_{1} + t_{2}}{2}$
B) $\frac{t_{1} - t_{2}}{2}$
C) $\sqrt{t_{1}t_{2}}$
D) $t_{1} + t_{2}$
Explanation
(c) $h = - vt_{1} + \frac{1}{2}gt_{1}^{2}$ or $\frac{h}{t_{1}} = - v + \frac{1}{2}gt_{1}$ …(i)
$h = vt_{2} + \frac{1}{2}gt_{1}^{2}$ or $- \frac{h}{t_{2}} = - v + \frac{1}{2}gt_{2}$ …(ii)
$\therefore\ \ \frac{h}{t_{1}} + \frac{h}{t_{2}} = \frac{1}{2}g(t_{1} + t_{2})$
or $h = \frac{1}{2}gt_{1}t_{2}$
For falls under gravity from the top of the tower
$h = \frac{1}{2}gt^{2}$
$\therefore\ \frac{1}{2}gt_{1}t_{2} = \frac{1}{2}gt^{2} \Rightarrow t = \sqrt{t_{1}t_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.