Question

A block of mass 1 Kg is dropped on a horizontal ground from a height of 128 m. If coefficient of restitution between ball and ground is 1/2, then height gained by ball after 3rd impact from ground is?

Answer 1

Initial height = 128m

coefficient of restitution(n) = $\frac{V_{after impact} }{V_{before impact}}$   (by definition)

Also coefficient of restitution(n) = √($\frac{h_{0} }{h_{1}}$)

where h₀ = initial height from which it is dropped
           h₁ = height upto which it goes after impact

After first impact, n = √($\frac{h_{0} }{h_{1}}$)

                             ⇒h₁ = h₀.n²

After second impact, n = √($\frac{h_{1} }{h_{2}}$)

                             ⇒h₂ = h₁.n² = h₀.n²*²

After third impact, n = √($\frac{h_{2} }{h_{3}}$)

                             ⇒h₃ = h₂.n² = h₀.n²*³

After mth impact, n = √($\frac{h_{m-1} }{h_{m}}$)

                             ⇒hm = h(m-₁).n² = h₀.(n^(2m))

Thus, for third impact, m=3
h₃ = h₀.n²*³ = 128.(1/2)⁶ = 128/64 = 2m