Question

If a body released from the top of a tower of height \(H\) meter takes \(T\) seconds to reach the ground, where is the body at time \(T/2\) seconds from the ground?

• A. \(\frac{H}{2}\)
• B. \(\frac{H}{4}\)
• C. \(\frac{3H}{4}\)
• D. \(\frac{2H}{3}\)

Hint:

For free fall from rest, distance fallen is proportional to \(t^2\). At half the total time, the body falls only one-fourth of the height.

Answer 1

The Correct Option is C

The body is released from rest from the top of a tower. So initial velocity is: \[ u=0. \] Let the height of the tower be: \[ H. \] The total time to reach the ground is: \[ T. \] Using the equation of motion: \[ s=ut+\frac{1}{2}gt^2. \] Since \(u=0\), \[ s=\frac{1}{2}gt^2. \] For the complete fall: \[ H=\frac{1}{2}gT^2. \] Now, at time: \[ t=\frac{T}{2}, \] distance fallen from the top is: \[ s=\frac{1}{2}g\left(\frac{T}{2}\right)^2. \] \[ s=\frac{1}{2}g\cdot \frac{T^2}{4}. \] \[ s=\frac{1}{8}gT^2. \] But from total height: \[ H=\frac{1}{2}gT^2. \] So, \[ \frac{1}{8}gT^2=\frac{1}{4}\left(\frac{1}{2}gT^2\right). \] \[ s=\frac{H}{4}. \] This is the distance fallen from the top. Therefore, the distance of the body from the ground is: \[ H-\frac{H}{4}. \] \[ =\frac{3H}{4}. \] Hence, at time \(T/2\), the body is: \[ \frac{3H}{4} \] above the ground.