Question:medium

A ball is dropped from a height of 20 m. What is its velocity just before hitting the ground? (Take \( g = 9.8 \, \text{m/s}^2 \))

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Remember: For free fall, use the equation \( v^2 = u^2 + 2gh \), where \( u = 0 \) for an object dropped from rest.
Updated On: Nov 26, 2025
  • \( 10 \, \text{m/s} \)
  • \( 14 \, \text{m/s} \)
  • \( 20 \, \text{m/s} \)
  • \( 18 \, \text{m/s} \)
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The Correct Option is B

Solution and Explanation

Step 1: Employ the equation of motion for velocity
The velocity of an object in free fall from a specific height is determined by:\[v^2 = u^2 + 2gh\]With the following variables:- \( v \) representing the final velocity,
- \( u \) denoting the initial velocity (zero for free fall),
- \( g \) indicating the acceleration due to gravity,
- \( h \) specifying the drop height.
Step 2: Insert the provided values Provided data:- Initial velocity \( u = 0 \, \text{m/s} \),
- \( g = 9.8 \, \text{m/s}^2 \),
- Height \( h = 20 \, \text{m} \).
Substituting these into the equation yields:\[v^2 = 0 + 2 \times 9.8 \times 20 = 392\]\[v = \sqrt{392} \approx 14 \, \text{m/s}\]Answer: The ball's velocity upon impact with the ground is approximately \( 14 \, \text{m/s} \). This corresponds to option (2).
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