Question:

When a body is dropped from height H, the time taken to reach the ground is T. What is the time taken to reach height H/2 from the ground?

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For free fall under gravity, the time taken to fall a certain fraction of the total height is proportional to the square root of that fraction.
Updated On: Apr 25, 2025
  • \( \frac{T}{2} \)
  • \( \frac{T}{\sqrt{2}} \)
  • \( \sqrt{2} T \)
  • \( \frac{T}{4} \)
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The Correct Option is B

Solution and Explanation


For a body falling freely under gravity, the equation of motion is given by: \[ h = \frac{1}{2} g t^2 \] Where \( h \) is the height, \( g \) is the acceleration due to gravity, and \( t \) is the time taken to fall a distance \( h \). - The time \( T \) taken to fall a height \( H \) is given by: \[ H = \frac{1}{2} g T^2 \quad \Rightarrow \quad T = \sqrt{\frac{2H}{g}} \] - The time \( t_{\frac{H}{2}} \) taken to fall to a height \( \frac{H}{2} \) is: \[ \frac{H}{2} = \frac{1}{2} g t_{\frac{H}{2}}^2 \quad \Rightarrow \quad t_{\frac{H}{2}} = \sqrt{\frac{H}{g}} = \frac{T}{\sqrt{2}} \] Thus, the time taken to reach half the height is \( \frac{T}{\sqrt{2}} \).
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