Question

A body released from the top of a tower of height H. It takes t time to reach the ground. The position of the body $\dfrac{t}{2}$ time after release will be
(A) at $\dfrac{H}{2}m$ from the ground
(B) at $\dfrac{H}{4}m$ from the ground
(C) at $\dfrac{{3H}}{4}m$ from the ground
(D) at $\dfrac{H}{6}m$ from the ground

Answer 1

The acceleration acting on the ball will be ‘g’ as it is just released from the top of the tower. Now apply the equation of motion to the values given. Now subtract the distance travelled from the total distance to find the distance from the ground.

Complete step by step answer:
The acceleration of the ball is g ,
Its initial velocity $U = 0$ ,
We know that distance s is ,
$s = ut + \dfrac{1}{2}a{t^2}$
But $u = 0$ , and acceleration is equal to g so,
$H = \dfrac{1}{2}g{t^2}$ $(u = 0)$
Therefore, in $\dfrac{t}{2}$ seconds,
$H' = \dfrac{1}{2}g{\left( {\dfrac{t}{2}} \right)^2}$
On solving the equation we get,
$H' = \dfrac{1}{2} \times \dfrac{{g{t^2}}}{4} = \dfrac{H}{4}$
Now subtracting the distance travelled in $\dfrac{t}{2}$ seconds from the total distance of the tower,
$H - \dfrac{h}{4} = \dfrac{{3H}}{4}$
Therefore, option (C) is correct.

Note:

The acceleration due to gravity is a force that pulls objects towards itself. Its value is $9.8m/{s^2}$ . It is denoted by the letter ‘g’. Always take acceleration due to gravity value as $9.8m/{s^2}$ unless it's told.
Any equation that describes the velocity of an object with respect to time is known as the equation of motion.
Initial velocity is the velocity from which a body starts to get accelerated.