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A 0.30 kg softball has a velocity v0 of 13.8 m/s at an angle of 35 degrees below the horizontal just before making contact with a bat. The ball leaves the bat 1.8 ms later with a vertical velocity of magnitude 10 m/s as shown in Fig. 10-28.

(a)What are the components of the average force on the softball during the collision?

(b)What is the magnitude of the average force of the bat on the ball during the ball-bat contact?

(c) What is the direction of the average force?
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x direction :

Fx = m (v2x - v1x) / t = 0.3 (13.8)(cos 35) / 0.0023 = 1474.47 N

Fy = m (v2y - v1y) / t = 0.3 ( 10 + 13.8 sin 35 ) / 0.0023 = 2336.78 N

so F = 1474.47 i + 2336.78 j N

b) so F = sqrt (Fx^2 + Fy^2) = 2763.08 N

c) theta = tan^-1 (Fy/Fx) = 57.7 degrees
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