# 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?

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Initial height = 128m

coefficient of restitution(n) = (by definition)

Also coefficient of restitution(n) = √()

where h₀ = initial height from which it is dropped

h₁ = height upto which it goes after impact

After__first__ impact, n = √()

⇒h₁ = h₀.n²

After__second__ impact, n = √()

⇒h₂ = h₁.n² = h₀.n²*²

After__third__ impact, n = √()

⇒h₃ = h₂.n² = h₀.n²*³

After__mth__ impact, n = √()

⇒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**

coefficient of restitution(n) = (by definition)

Also coefficient of restitution(n) = √

where h₀ = initial height from which it is dropped

h₁ = height upto which it goes after impact

After

⇒h₁ = h₀.n²

After

⇒h₂ = h₁.n² = h₀.n²*²

After

⇒h₃ = h₂.n² = h₀.n²*³

After

⇒hm = h(m-₁).n² = h₀.(n^(2m))

Thus, for third impact, m=3

h₃ = h₀.n²*³ = 128.(1/2)⁶ = 128/64 =