Golf Ball Impact Force: Average Calculated

[LOannie234]

A golf ball (m = 71.7 g) is struck a blow that
makes an angle of 43.1
with the horizontal.
The drive lands 122 m away on a flat fairway.
The acceleration of gravity is 9.8 m/s^2.

If the golf club and ball are in contact for
6.62 ms, what is the average force of impact?
Neglect air resistance

I've tried using t=x/vcos(theta) and plugged it into y=yo+vt-.5gt^2 and used f=mv/t

 

[Sniperman724]

you could try to use ft=p where p equals the change in momentum, and f equals the force applied to the ball. You could try to work backwards using kinematics equations to find how fast the ball will be going at the end of its trajectory, then because we neglect air resistance, the ball is traveling the same horizontal speed at both its starting and ending points (since it is a parabolic arc)

 

[LOannie234]

So for the horizontal velocity, would the equation be

v=(square root of)mg/2cos(theta)?

 

[Sniperman724]

If the ball's initial velocity is at an angle of 43.1 degrees and it lands on a flat fairway, then the ball lands at 43.1 degrees as well.
I am not sure if that helps at all

 

[LOannie234]

Then

v=(square root)9.8m/s^2*122m/2cos(43.1)
v=20.8924m/s

If so

x=vt
t=x/v
t=5.839s

Am I following the correct path?

 

[Sniperman724]

Yea, I would think so, so now that you have the velocity of the ball in the air, you can find the ball's initial acceleration, then plug that into f=ma to find the force that was needed

I really hope this helps