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 th

Question

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?

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Thanh Hà 3 years 2021-07-22T04:07:43+00:00 1 Answers 140 views 0

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  1. ngochoa
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    2021-07-22T04:09:34+00:00
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    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.

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