# After falling from rest to a height of 30m, a 0.50kg ball rebounds upward, reaching a height of 20m. If the contact between ball and ground lasted 2.0m/s. What average force was exerted on the ball?

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Expert Answers

gsenviro | Certified Educator

Mass of the ball, m = 0.5 kg

Initial height = 30 m

Final height = 20 m

Contact time, t = 2 ms = 2 x 10^-3 s

Velocity during descent, V1 = `sqrt(2gh) = sqrt(2xx9.81xx30) = 24.3 m/s`

Velocity during ascent, V2 = `sqrt(2gh) = sqrt(2xx9.81xx20) = 19.8 m/s`

Momentum change during the impact = m(V1-V2) = 0.5 (24.3-(-19.8)) kg m/s

= 0.5 x 44.1 kg m/s = 22.05 kg m/s

(note that downward velocity is assumed positive and upward as negative and hence the negative sign)

Now, force is rate of change of momentum.

Thus, force exerted by the ball = momentum change/impact duration

= 22.05/(2x10^-3) N = **11025 N**.

Hope this helps