Question

A ball of mass 0.2 kg is thrown vertically down from a height of 10 m. It collides with the floor and loses 50% of its energy and then rises back to the same height. The value of its initial velocity is

• A. 14 ms^-1
• B. 196 ms^-1
• C. 20 ms^-1
• D. Zero

Answer 1

The Correct Option is A

Given: 

• Mass of the ball, \( m = 0.2 \) kg
• Initial height, \( h = 10 \) m
• Energy lost after collision = 50%
• Acceleration due to gravity, \( g = 9.8 \) m/s²

Step 1: Use Energy Conservation

Let initial velocity be \( u \). The total mechanical energy before hitting the ground:

\[ E_i = \frac{1}{2} m u^2 + mgh \]

After hitting the ground, the ball loses 50% of its energy and rises back to the same height, meaning its remaining energy is:

\[ E_f = \frac{1}{2} E_i \]

Step 2: Express in Terms of Initial Velocity

Since the ball reaches the same height after losing 50% energy, the energy left is equal to its potential energy at height \( h \):

\[ mgh = \frac{1}{2} \left( \frac{1}{2} m u^2 + mgh \right) \]

Solving for \( u \):

\[ 2mgh = \frac{1}{2} m u^2 + mgh \]

\[ mgh = \frac{1}{2} m u^2 \]

\[ u^2 = 2gh \]

\[ u = \sqrt{2 \times 9.8 \times 10} \]

\[ u = \sqrt{196} = 14 \text{ m/s} \]

Answer: The initial velocity of the ball is 14 m/s (Option A).