Q. A body is released from the top of a tower of height $h$ metre. It takes $T$ second to reach the ground. Where is the ball at the time $\frac{T}{2}$ second?
A) At $\frac{h}{4}$ metre from the ground
B) At $\frac{h}{2}$ metre from the ground
C) At $\frac{3h}{4}$ metre from the ground
D) Depends upon the mass and volume of the ball
Explanation
(c) $h = \frac{1}{2}gT^{2},h^{'} = \frac{1}{2}g\left( \frac{T}{2} \right)^{2}h^{'} = \frac{h}{4}$,
Height above the ground $= h - \frac{h}{4} = \frac{3h}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.