Question:
An object when dropped from a height \(h\) from the ground, reaches the ground in \(t\) s. The time after which the object was passing through a point at a height \(h/2\) from the ground is
An object when dropped from a height \(h\) from the ground, reaches the ground in \(t\) s. The time after which the object was passing through a point at a height \(h/2\) from the ground is
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Time in free fall varies as square root of height.
Updated On: Apr 24, 2026
- \(\sqrt{2}t\)
- \(\frac{t}{\sqrt{2}}\)
- \(\frac{t}{2}\)
- \(2t\)
- \(\frac{t}{4}\)
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The Correct Option is B
Solution and Explanation
Concept:
Equation of motion:
\[
h = \frac{1}{2}gt^2
\]
Step 1: Total fall time.
\[ h = \frac{1}{2}gt^2 \]
Step 2: For height \(h/2\).
\[ \frac{h}{2} = \frac{1}{2}g t_1^2 \] Substitute \(h = \frac{1}{2}gt^2\): \[ \frac{1}{4}gt^2 = \frac{1}{2}g t_1^2 \]
Step 3: Solve for \(t_1\).
\[ t_1 = \frac{t}{\sqrt{2}} \]
Step 1: Total fall time.
\[ h = \frac{1}{2}gt^2 \]
Step 2: For height \(h/2\).
\[ \frac{h}{2} = \frac{1}{2}g t_1^2 \] Substitute \(h = \frac{1}{2}gt^2\): \[ \frac{1}{4}gt^2 = \frac{1}{2}g t_1^2 \]
Step 3: Solve for \(t_1\).
\[ t_1 = \frac{t}{\sqrt{2}} \]
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