Question:
A ball is thrown upward from the ground with an initial speed of \(V\). At the same instant, another ball is dropped from a building of height \(20\,\text{m}\). If the balls are at the same height after \(0.8\,\text{s}\), then the magnitude of \(V\) is
A ball is thrown upward from the ground with an initial speed of \(V\). At the same instant, another ball is dropped from a building of height \(20\,\text{m}\). If the balls are at the same height after \(0.8\,\text{s}\), then the magnitude of \(V\) is
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When two bodies move under the same acceleration due to gravity, the common \(-\frac{1}{2}gt^2\) term cancels while comparing their heights at the same time.
Updated On: Jun 26, 2026
- \(15\,\text{ms}^{-1}\)
- \(25\,\text{ms}^{-1}\)
- \(12.5\,\text{ms}^{-1}\)
- \(18.5\,\text{ms}^{-1}\)
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The Correct Option is B
Solution and Explanation
Step 1: Write the height of the ball thrown upward.
For the ball thrown upward from the ground, \[ u=V,\quad a=-g \] The height after time \(t\) is \[ h_1=Vt-\frac{1}{2}gt^2 \]
Step 2: Write the height of the ball dropped from the building.
The second ball is dropped from height \[ 20\,\text{m} \] Its initial velocity is \[ 0 \] So, its height from the ground after time \(t\) is \[ h_2=20-\frac{1}{2}gt^2 \]
Step 3: Use the condition that both balls are at the same height.
Given, \[ t=0.8\,\text{s} \] At the same height, \[ h_1=h_2 \] So, \[ Vt-\frac{1}{2}gt^2=20-\frac{1}{2}gt^2 \] Cancel \[ -\frac{1}{2}gt^2 \] from both sides: \[ Vt=20 \] \[ V=\frac{20}{t} \]
Step 4: Substitute \(t=0.8\).
\[ V=\frac{20}{0.8} \] \[ V=25 \] Thus, \[ V=25\,\text{ms}^{-1} \]
Step 5: Final conclusion.
Hence, the magnitude of \(V\) is \[ \boxed{25\,\text{ms}^{-1}} \]
For the ball thrown upward from the ground, \[ u=V,\quad a=-g \] The height after time \(t\) is \[ h_1=Vt-\frac{1}{2}gt^2 \]
Step 2: Write the height of the ball dropped from the building.
The second ball is dropped from height \[ 20\,\text{m} \] Its initial velocity is \[ 0 \] So, its height from the ground after time \(t\) is \[ h_2=20-\frac{1}{2}gt^2 \]
Step 3: Use the condition that both balls are at the same height.
Given, \[ t=0.8\,\text{s} \] At the same height, \[ h_1=h_2 \] So, \[ Vt-\frac{1}{2}gt^2=20-\frac{1}{2}gt^2 \] Cancel \[ -\frac{1}{2}gt^2 \] from both sides: \[ Vt=20 \] \[ V=\frac{20}{t} \]
Step 4: Substitute \(t=0.8\).
\[ V=\frac{20}{0.8} \] \[ V=25 \] Thus, \[ V=25\,\text{ms}^{-1} \]
Step 5: Final conclusion.
Hence, the magnitude of \(V\) is \[ \boxed{25\,\text{ms}^{-1}} \]
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