Given Information:
mass of ball = 0.45 kg
Initial height = h₀ = 1 m
Initial velocity = v₀ = 12 m/s
coefficient of restitution = e = 0.6
Required Information:
height after bouncing back = h₂ = ?
Answer:
height after bouncing back = h₂ ≈ 3 m
Explanation:
We know from the conservation of energy principle
KE₀ + PE₀ = KE₁ + PE₁
½mv₀² + mgh₀ = ½mv₁² + mgh₁
Where v₁ is the velocity when ball strikes the ground and h₁ is the height at that instant which is zero.
½*0.45*(12)² + 0.45*9.81*1 = ½*0.45*v₁² + 0.45*9.81*0
32.4 + 4.414 = 0.225v₁² + 0
0.225v₁² = 36.81
v₁² = 36.81/0.225
v₁² = 163.6
v₁ = √163.6
v₁ = 12.79 m/s
The coefficient of restitution at the instant of impact is given by
e = v₁'/v₁
Where v₁' is the velocity of ball after the impact
v₁' = ev₁
v₁' = 0.6*12.79
v₁' = 7.674 m/s
The energy relation after the impact is
KE₁ + PE₁ = KE₂ + PE₂
½mv₁'² + mgh₁ = ½mv₂² + mgh₂
Where h₂ is the height of ball after the bounce and h₁ and v₂ are zero
½*0.45*(7.674)² + 0.45*9.81*(0) = ½*0.45*(0)² + 0.45*9.81*h₂
½*0.45*(7.674)² + 0 = 0 + 0.45*9.81*h₂
13.25 = 4.414h₂
h₂ = 13.25/4.414
h₂ = 3.001 m
h₂ ≈ 3 m
Therefore, the ball will reach a height of 3 m above the ground on its first bounce.