Question:medium

A ball is thrown upwards with a velocity of \(20\,m/s\). Find the maximum height reached \((g=10\,m/s^2)\).

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For vertical upward motion, maximum height can also be found directly using: \[ h = \frac{u^2}{2g} \] This shortcut is very useful in projectile and vertical motion problems.
Updated On: Apr 18, 2026
  • \(10\,m\)
  • \(20\,m\)
  • \(30\,m\)
  • \(40\,m\)
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The Correct Option is B

Solution and Explanation

Topic: Kinematics - Motion under Gravity
Step 1: Understanding the Question:
When an object is thrown upwards, it decelerates due to gravity until its velocity reaches zero at the peak.
We need to find this peak distance (maximum height).
Step 2: Key Formula or Approach:
Use the kinematic equation:
\[ v^2 = u^2 - 2gh \]
At maximum height, the final velocity \(v = 0\).
Step 3: Detailed Explanation:
1. List the known variables:
Initial velocity \(u = 20 \, \text{m/s}\).
Acceleration due to gravity \(g = 10 \, \text{m/s}^2\).
Final velocity at peak \(v = 0\).
2. Rearrange the formula to solve for \(h\):
\[ 0 = u^2 - 2gh \implies h = \frac{u^2}{2g} \]
3. Substitute the values:
\[ h = \frac{20^2}{2 \times 10} = \frac{400}{20} \]
\[ h = 20 \, \text{m} \]
Step 4: Final Answer:
The maximum height reached is \(20 \, \text{m}\).
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