Question

Derive the expression of maximum height for a ball after n collisions if it loses 3/4 of its energy after each collision when dropped from a height h to a concrete floor

Answer 1

When the ball is at a height h, its total energy = E₀
       = Potential energy + kinetic energy = m g h + 1/2 m 0² = mg h
                 where m = mass of the ball and g is the acceleration due to gravity.

When the ball reaches the floor, its energy = PE + KE = 0 + KE = KE₀
     Because energy is conserved, KE₀ = E₀ = m g h 

When the ball bounces back, 3/4 E₀ is lost and only 1/4 E₀ or 1/4 mgh remains.

After rebounding (n =1) the ball reaches a height = h/4, at which its PE = 1/4 mgh
                      E₁ = 1/4 m g h              h₁ = h/4

Next collision (n=2) the energy before collision is E₁ and after = E₂ = 1/4 E₁
                        E₂ = 1/4 [ 1/4 mg h] = 1/4²  m g h 

After n th collision with the floor, the ball attains a height of 
$\ \ \ \ \ \ E_n = (\frac{1}{4})^n E_0.\\.\ \ \ \ \ h_n = (\frac{1}{4})^n h=\frac{h}{4^n}$

Answer 2

Initial : PE = mgh , KE = 0 , So total energy = mgh
Just before collision : PE=0, KE= mgh-0=mgh
Just after collision :KE = mgh/4 =TE
So maximum height reached = h/4
You can see that only one-fourth of energy is carried from one collision to another.
So after n-collisions, height = h/(4^n)

Coefficient of restitution(n) = velocity before collision/velocity after collision
                                                = √h₂/h₁
                                               = √(h/4)/h
                                         ⇒ n = 0.5