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Question:

A ball is thrown vertically downwards from a height of 20 m with an initial velocity υ₀. It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity υ₀ is (Take g=10ms⁻²)

10ms⁻¹

20ms⁻¹

28ms⁻¹

14ms⁻¹

Solution:

Initial energy Eᵢ = PE + KE = mgh + 1/2mv²
Energy when it reaches the ground: E = 1/2mv'² since PE is zero.
Given, E = 1/2Eᵢ
Since it rises to the same height Ef = mgh + 0 is the final energy
⇒1/2(mgh + 1/2mv²) = Ef = mgh + 0
v² = 2gh = 2 × 10 × 20 = 400
⇒v = 20m/s