Question

A body is released from the top of a tower of height h. It takes t sec to reach the ground. Where will be the ball after time $\frac{t}{2}$ s.

• A. At h/2 from the ground
• B. At h/4 from the ground
• C. Depends upon mass and volume of the body
• D. At 3h/4 from the ground

Answer 1

The correct option is

B

At 3h/4 from the ground

Let the body after time $\frac{t}{2}$ fall a distance x from the top. Since it starts from rest i.e. initial velocity u=0, here a will be acceleration due to gravity i.e. a=g

using eq. of motion: $s=ut+\frac{1}{2}a{t}^2$ ,we have

$x=\frac{1}{2}g\frac{t^2}{4}=\frac{g{t}^2}{8}$ .............(i)

It reaches the ground in t seconds, we have

$h=\frac{1}{2}g{t}^2$ ..............(ii)

Eliminate t from (i) and (ii), we get x = $\frac{h}{4}$

$\therefore$ Height of the body from the ground

= $h-\frac{h}{4}=\frac{3h}{4}$