Question

A golf ball (m = 46.0 g) is struck with a force that makes an angle of 44.4° with the horizontal. the ball lands 193 m away on a flat fairway. if the golf club and ball are in contact for 6.92 ms, what is the average force of impact? (neglect air resistance.)

Answer 1

Using the formula for bodies in projectile motion, we can calculate for the velocity v:

 d = (v^2) * (sin2theta) / g 

Substituting the given values:

193 m = (v^2) * (sin(2*44.4)) / (9.8 m/s^2)

v= 50.5 m/s

 

Calculating for the impulse I,

I = mf * vf – mi * vi

since mf = mi and vi = 0 (the golf ball starts form rest)

I = m * vf

I = 0.046 kg * 50.5 m/s

I = 2.323 kg m/s

 

Calculating for the force of impact:

I = F t

F = (2.323 kg m/s) / (6.92 x 10^-3 s)

F =335.7 kg m/s^2 = 335.7 N