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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) =  \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

1 5 1
clearly understood
second formula for restitution is important, try to derive it. I have directly given it.
Then my efforts have some value. I am glad.
it was explained in my book too but i was not able to understand it so thanks
welcome buddy!!
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