Question

A ball is thrown vertically downwards from a height of 20 m with an initial velocity v₀. It collides with the ground loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity v₀ is : (Take g = 10 ms⁻²)(a) 20 ms⁻¹(b) 28 ms⁻¹(c) 10 ms⁻¹(d) 14 ms⁻¹

Answer 1

Let v is the velocity of ball after collision to reach the same height e.g., h = 20m

at height, h = 20 m, velocity of it will be zero.

so, v² = 0 + 2gh = 2 × 10 × 20 = 400

v = 20 m/s

now, energy of body after collision = 1/2 mv²

= 1/2 m(20)² = 200m

initial energy of body = kinetic energy+ potential energy

= 1/2 mv₀² + mgh

a/c to question,

final energy = 50% of initial energy

200m = 1/2 × Initial energy

so, Initial energy = 400m

now, 1/2 mv₀² + mgh = 400m

or, 1/2v₀² + gh = 400

or, 1/2 v₀² + 10 × 20 = 400

or, v₀² = 400 => v₀ = 20 m/s

hence, option (a) is correct.

Answer 2

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