A 1kg ball is dropped of a 20m tall bridge onto tarmac. The ball experiences 2N of drag throughout its motion. The ground has a coefficient of restitution of 0.5. What is the maximum height the ball will reach after one bounce
Energy
GPE of the ball on the bridge: MGH = 1 * 9.81 * 20 =196 J
energy lost to drag = F * D = 2 * 20 = 40 J
Velocity of the ball just before impact =sqr(2*(196-40)/1) =17.66 m/s
Velocity of the ball just after impact = 17.660.5 = 8.83 m/s
Kinetic energy just after impact 0.5 * 1 * 8.83^2 =39 J
Equation for final height (using energy)
KE = Drag energy + GPE
39 = 2H + 19.81H
H=3.30m
MD
Answered by Max D. • Further Mathematics tutor
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