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?
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?
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For free fall from rest, distance fallen is proportional to \(t^2\).
Updated On: May 7, 2026
- \(\frac{H}{2}\)
- \(\frac{H}{4}\)
- \(\frac{3H}{4}\)
- \(\frac{2H}{3}\)
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The Correct Option is C
Solution and Explanation
Concept:
For a freely falling body released from rest: \[ s=\frac{1}{2}gt^2 \]
Step 1: Total height of tower is \(H\), and total time taken is \(T\). \[ H=\frac{1}{2}gT^2 \]
Step 2: Distance fallen in time \(T/2\) is: \[ s=\frac{1}{2}g\left(\frac{T}{2}\right)^2 \] \[ s=\frac{1}{2}g\frac{T^2}{4} \] \[ s=\frac{1}{4}\left(\frac{1}{2}gT^2\right) \] \[ s=\frac{H}{4} \]
Step 3: This is the distance fallen from the top.
Step 4: Height from the ground at that instant is: \[ H-\frac{H}{4} \] \[ =\frac{3H}{4} \] \[ \boxed{\frac{3H}{4}} \]
For a freely falling body released from rest: \[ s=\frac{1}{2}gt^2 \]
Step 1: Total height of tower is \(H\), and total time taken is \(T\). \[ H=\frac{1}{2}gT^2 \]
Step 2: Distance fallen in time \(T/2\) is: \[ s=\frac{1}{2}g\left(\frac{T}{2}\right)^2 \] \[ s=\frac{1}{2}g\frac{T^2}{4} \] \[ s=\frac{1}{4}\left(\frac{1}{2}gT^2\right) \] \[ s=\frac{H}{4} \]
Step 3: This is the distance fallen from the top.
Step 4: Height from the ground at that instant is: \[ H-\frac{H}{4} \] \[ =\frac{3H}{4} \] \[ \boxed{\frac{3H}{4}} \]
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