Thank you for reporting, we will resolve it shortly
Q.
A body is released from the top of the tower H metre high. It takes t second to reach the ground. Where is the body after t/2 second of release ?
Motion in a Straight Line
Solution:
Applying $S = ut + \frac{1}{2} gt^2 $ for the Ist case
$H = \frac{1}{2} gt^2$ .....(i)
Let $H_1$ be the height after t/2 secs. So distance of fall $= H - H_1$
$H - H_1 = \frac{1}{2} g \left( \frac{t}{2} \right)^2$
$\Rightarrow \, H - H_1 = \frac{1}{8} gt^2$ ....(ii)
Dividing (i) and (ii),
$\frac{H - H_1}{H} = \frac{1}{8} \times \frac{2}{1} = \frac{1}{4}$
$\Rightarrow \, 4H - 4H_1 = H \, \Rightarrow \, H_1 = \frac{3}{4} H $