A 60 g golf ball leaves the face of a golf club with a velocity of 75 m/s. If the club exerted an average force of 3.0 x 10^4 N, what was the time of impact between the club and the ball?

Answer 1

f=m*a

30000N = 60g * a

a = (change in velocity) / (change in time)

30000N = 60g * (Vf - Vi) / (change in time)

500 = (75 - 0) / (change in time)

change in time = .15 sec

the change in time is the time that the ball was in contact with the club

Answer 2

the mass you use is not 60g, it is .06kg. Sorry. Answer is .00015 sec.