Question

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 ?

• A. at 3H/4 metre from the ground
• B. at H/2 metre from the ground
• C. at H/6 metre from the ground
• D. at H/4 metre from the ground

Answer 1

Answer: (A)

at 3H/4 metre from the ground

Answer 2

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), $\frac{H - H_1}{H} = \frac{1}{8} \times \frac{2}{1} = \frac{1}{4}$ $\Rightarrow \, 4H - 4H_1 = H \, \Rightarrow \, H_1 = \frac{3}{4} H$