An object is dropped from a height h from the ground. Every time it hits the ground it loses 50 % of its kinetic energy. The total distance covered as t → ∞ is:

Step-by-step solution

Since the object is dropped from height, thus its potential energy is equal to its kinetic energy. As per question, kinetic energy after hitting the ground is 1 2 m v ′ 2 = 1 2 × 1 2   m v 2 v ′ = v 2 v = e u       e = 1 2 H = h 1 + e 2 1 - e 2 The total distance covered = h 1 + 1 2 1 - 1 2 = 3 h