The 0.45-kg soccer bal is 1 m above the ground when it is kicked upward at 12 m/s. If the coefficient of restitution between the ball and the ground is e = 0.6, what maximum height above the ground does the ball reach on its first bounce?

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.