A 0.140-kg baseball is pitched horizontally at 36.7 m/s. When a player hits the ball, it moves at the same speed, but in the opposite direction. If the bat and the ball are in contact for 0.450 ms, calculate the average force the bat exerts on the ball.

Question

Work Shown:
F/\t0.450ms=(0.140kg*36.7m/s)(0.140kg*-36.7m/s)
.450*1000=450s
F/\450s=(0.140kg*36.7m/s)(0.140kg*-36.7m/s)
F/\450s=(5.138kg*m/s)*(-5.138kg*m/s)
F/\450s=-26.39904kg*m/s
(26.39904kg*m/s)/(/\t450s)=0.0586645kg*m

I've asked my teacher for help, but the solution we both get doesn't equal the one in the solutions manual. Please Help!

P.S. /\t is change in time. They don't have the delta symbol

Answer 1

0.450 ms = 0.000450 s.

V = Vo + a*t = -36.7
36.7 + a*0.000450 = -36.7
0.00045a = -36.7-36.7 = -73.4
a = -163,111 m/s^2

F = M*a = 0.140 * (-163111) = 22,836 N.

Answer 2

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