Question:medium

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

Show Hint

At the highest point of vertical motion, the ball is momentarily at rest. Think about how the initial kinetic energy at launch relates to the gravitational potential energy gained by the time it stops rising, and set the two equal to connect speed, gravity and height.
Updated On: Aug 17, 2026
  • \(10\,m\)
  • \(20\,m\)
  • \(30\,m\)
  • \(40\,m\)
Show Solution

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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