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)

Applying $S=ut+\frac{1}{2}g{t}^2$ for the Ist case
$H=\frac{1}{2}g{t}^2$ .....(i)
Let $H_1$ be the height after t/2 secs. So distance of fall $=H-H_1$
$H-H_1=\frac{1}{2}g{\left(\frac{t}{2}\right)}^2$
$\Rightarrow H-H_1=\frac{1}{8}g{t}^2$ ....(ii)
Dividing (i) and (ii),
$\frac{H-H_1}{H}=\frac{1}{8}\times \frac{2}{1}=\frac{1}{4}$
$\Rightarrow 4H-4H_1=H\Rightarrow H_1=\frac{3}{4}H$