Energy of Rebound – Ideal Elastic Bounce A rubber ball is dropped from a height of 2 m. Assuming no loss of velocity on rebound (perfectly elastic impact with the floor), to what maximum height will the ball rise?

Introduction / Context:
When a ball rebounds without energy loss (coefficient of restitution e = 1), mechanical energy is conserved between just-before-impact and just-after-rebound speeds, leading to symmetric rise and fall heights (ignoring air resistance).

Given Data / Assumptions:

• Initial drop height h = 2 m.
• Perfectly elastic impact with the floor (e = 1).
• Neglect air drag and rolling effects.

Concept / Approach:
Use energy conservation or kinematics: the speed just before impact equals the speed just after rebound (upward) if e = 1. Therefore, the ball will climb back to the same height from which it was dropped.

Step-by-Step Solution:
Before impact: v_down = sqrt(2 g h). Elastic rebound: v_up = v_down. As it rises: v_up^2 = 2 g h′ ⇒ h′ = v_up^2 / (2 g) = (2 g h) / (2 g) = h = 2 m.

Verification / Alternative check:
Using coefficient of restitution: e = v_after / v_before = 1 ⇒ equal speeds; hence equal heights.

Why Other Options Are Wrong:
1 m, 3 m, 4 m contradict energy conservation for e = 1; without losses the ball returns to its original height. “It will not rise” is obviously false for a bounce.

Common Pitfalls:
Assuming some loss automatically; here the problem explicitly states no loss.

Final Answer:
2 m