# Physics

After a 0.200-kg rubber ball is dropped from a height of 1.95 m, it bounces off a concrete floor and rebounds to a height of 1.45 m.

(a) Determine the magnitude and direction of the impulse delivered to the ball by the floor.

(b) Estimate the time the ball is in contact with the floor to be 0.08 seconds. Calculate the average force the floor exerts on the ball.

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1. a. V^2 = Vo^2 + 2g*h = 0 @ max ht.
Vo^2 - 19.6*1.45 = 0
Vo^2 = 19.6*1.45 = 28.42
Vo = 5.33 m/s = Initial velocity after
striking the floor.

Impulse = m*V = 0.20*5.33 = 1.067 kg-m/s

b. V^2 = Vo^2 + 2g*h = 0 + 19.6*1.95 =
38.22
V = 6.18 m/s. = Velocity just before
striking the floor.

a = (Vo-V)/t =
(-5.33-6.18)/0.08 = -143.9 m/s^2

F = m*a = 0.20 * (-143.9) = -28.78 N.

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