Asked by alika

A golf club strikes a 0.040-kg golf ball in order to launch it from the tee. For simplicity, assume that the average net force applied to the ball acts parallel to the ball’s motion, has a magnitude of 7950 N, and is in contact with the ball for a distance of 0.011 m. With what speed does the ball leave the club?
Answered by Henry
a = F/m = 7950/0.040 = 198,750 m/s^2.

V^2 = Vo^2 + 2a*d
Vo^2 = V^2 - 2a*d = 0 + 2*198750*0.011 =
4373
Vo = 66.1 m/s.
Answered by Elizabeth
Am I the only one seeing this in 2021, when this was posted in 2014?????
Answered by conner
bro got ratio'd on jishka
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