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A golf club strikes a 0.045-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 6800 N, and is in contact with the ball for a distance of 0.010 m. With what speed does the ball leave the club?

Respuesta :

Answer:

speed of ball is 54.97 m/s

Explanation:

given data

mass of ball m = 0.045 kg

distance s = 0.010 m

net force F  = 6800 N

to find out

what speed does the ball leave the club

solution

we consider W = change in kinetic energy

so F×s = 1/2 mv²

we will find v  from here

v = [tex]\sqrt{(2F*s/m)}[/tex]

put all value

v =  [tex]\sqrt{(2*6800*0.010/0.045)}[/tex]

v = 54.9747 m/s

speed of ball is 54.97 m/s

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