Question:
A ball is dropped from a tower of height \(80\,\text{m}\). The time it takes to cover the last \(50\%\) of its fall is (acceleration due to gravity \(=10\,\text{m s}^{-2}\))
A ball is dropped from a tower of height \(80\,\text{m}\). The time it takes to cover the last \(50\%\) of its fall is (acceleration due to gravity \(=10\,\text{m s}^{-2}\))
Show Hint
In free fall from rest,
\[
s=\frac{1}{2}gt^2.
\]
To find the time taken for a specific part of the journey, first find the times corresponding to the starting and ending positions and then subtract them.
Updated On: Jun 18, 2026
- \(2\sqrt{2}\,\text{s}\)
- \(1.17\,\text{s}\)
- \(4\,\text{s}\)
- \(2.0\,\text{s}\)
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The Correct Option is B
Solution and Explanation
Step 1: Find the total time of fall.
The height of the tower is \[ h=80\,\text{m} \] Since the ball is dropped from rest, \[ h=\frac{1}{2}gt^2 \] \[ 80=\frac{1}{2}(10)t^2 \] \[ 80=5t^2 \] \[ t^2=16 \] \[ t=4\,\text{s} \] Thus, the total time of fall is \[ 4\,\text{s}. \]
Step 2: Find the time taken to fall the first \(40\,\text{m}\).
The last \(50\%\) of the fall corresponds to the last \[ 40\,\text{m} \] of the total \[ 80\,\text{m}. \] For a displacement of \[ 40\,\text{m}, \] \[ 40=\frac{1}{2}(10)t_1^2 \] \[ 40=5t_1^2 \] \[ t_1^2=8 \] \[ t_1=2\sqrt{2}\,\text{s} \]
Step 3: Determine the time for the last \(40\,\text{m}\).
The ball reaches the halfway point at time \[ t_1=2\sqrt{2}\,\text{s}. \] Hence, the time spent in covering the last \(40\,\text{m}\) is \[ t_{\text{last}} = 4-2\sqrt{2} \] \[ =4-2(1.414) \] \[ =4-2.828 \] \[ =1.172\,\text{s} \] \[ \approx 1.17\,\text{s} \]
Step 4: Final conclusion.
Therefore, the time required to cover the last \(50\%\) of the fall is \[ \boxed{1.17\,\text{s}} \]
The height of the tower is \[ h=80\,\text{m} \] Since the ball is dropped from rest, \[ h=\frac{1}{2}gt^2 \] \[ 80=\frac{1}{2}(10)t^2 \] \[ 80=5t^2 \] \[ t^2=16 \] \[ t=4\,\text{s} \] Thus, the total time of fall is \[ 4\,\text{s}. \]
Step 2: Find the time taken to fall the first \(40\,\text{m}\).
The last \(50\%\) of the fall corresponds to the last \[ 40\,\text{m} \] of the total \[ 80\,\text{m}. \] For a displacement of \[ 40\,\text{m}, \] \[ 40=\frac{1}{2}(10)t_1^2 \] \[ 40=5t_1^2 \] \[ t_1^2=8 \] \[ t_1=2\sqrt{2}\,\text{s} \]
Step 3: Determine the time for the last \(40\,\text{m}\).
The ball reaches the halfway point at time \[ t_1=2\sqrt{2}\,\text{s}. \] Hence, the time spent in covering the last \(40\,\text{m}\) is \[ t_{\text{last}} = 4-2\sqrt{2} \] \[ =4-2(1.414) \] \[ =4-2.828 \] \[ =1.172\,\text{s} \] \[ \approx 1.17\,\text{s} \]
Step 4: Final conclusion.
Therefore, the time required to cover the last \(50\%\) of the fall is \[ \boxed{1.17\,\text{s}} \]
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