Question

A ball of mass 0.1 kg, initially at rest, is dropped from height of 1 m. Ball hits the ground and bounces off the ground. Upon impact with the ground, the velocity reduces by 20%. The height (in m) to which the ball will rise is:

• A. \(0.64\)
• B. \(0.86\)
• C. \(0.34\)
• D. \(0.72\)

Hint:

For any bouncing problem where velocity is scaled by a factor \(e\) (the coefficient of restitution), the height scales as the square of that factor: \[ h_{\text{new}} = e^2 \times h_{\text{initial}} \] Here, \(e = 0.8\), so \(h_{\text{new}} = (0.8)^2 \times 1 = 0.64 \text{ m}\). Notice that the mass of the ball (0.1 kg) does not affect the kinematic motion.

Answer 1

The Correct Option is A

Concept: When an object is dropped under the influence of uniform gravity from an initial height \(h_1\), its potential energy converts completely into kinetic energy just before impact. Upon bouncing, it leaves the ground with a reduced velocity, and its remaining kinetic energy converts back into potential energy as it rises to a new maximum height \(h_2\).

Step 1: Calculating the velocity of the ball just before hitting the ground.
Let \(h_1 = 1 \text{ m}\) be the initial drop height, and let \(g\) be the acceleration due to gravity. Using the third equation of motion (\(v^2 = u^2 + 2as\)) with an initial downward velocity \(u = 0\): \[ v_1^2 = 0 + 2gh_1 \quad \Rightarrow \quad v_1 = \sqrt{2gh_1} \] Substituting \(h_1 = 1\): \[ v_1 = \sqrt{2g(1)} = \sqrt{2g} \]

Step 2: Calculating the rebound velocity after ground impact.
The problem states that upon hitting the ground, the ball's velocity is reduced by 20%. This means it retains 80% of its approach velocity. Let \(v_2\) be the upward rebound velocity immediately after the bounce: \[ v_2 = v_1 - (20% \text{ of } v_1) = 80% \text{ of } v_1 \] \[ v_2 = 0.80 \times v_1 \] Substituting our expression for \(v_1\) from
Step 1: \[ v_2 = 0.8 \sqrt{2g} \]

Step 3: Computing the subsequent rebound height \(h_2\).
Let \(h_2\) be the maximum height the ball reaches after bouncing. At this peak height, its upward velocity drops to zero. Applying the equation of motion for this upward phase: \[ 0 = v_2^2 - 2gh_2 \quad \Rightarrow \quad 2gh_2 = v_2^2 \] Solving for \(h_2\): \[ h_2 = \frac{v_2^2}{2g} \] Now substitute our expression for \(v_2 = 0.8\sqrt{2g}\) into this formula: \[ h_2 = \frac{(0.8 \sqrt{2g})^2}{2g} \] Squaring the terms in the numerator: \[ h_2 = \frac{(0.8)^2 \times (2g)}{2g} \] Canceling the common factor \(2g\) from both the numerator and denominator: \[ h_2 = (0.8)^2 = 0.64 \text{ m} \] Thus, the ball will rise to a height of 0.64 meters, which corresponds to option (1).