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