Calculating Impulse of Golf Ball - Average Force of Impact

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The discussion focuses on calculating the average force of impact when a golf ball is struck at a 45-degree angle, landing 200 meters away. The calculations show that the initial velocity of the ball is approximately 44.3 m/s, leading to an impulse of 2.04 Ns. Using the impulse-momentum theorem, the average force of impact is determined to be around 291.4 N, based on a contact time of 7 milliseconds. The method used for the calculations is confirmed to be correct, though the distance achieved with a pitching wedge at that angle raises some skepticism. Overall, the calculations provide a solid understanding of the dynamics involved in the impact of a golf ball.
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A golf ball (m=46g) is struck a blow that makes an angle of 45 degrees with the horizontal. The ball lands 200 m away on a flat fairway. If the golf club and ball are in contact for 7 milliseconds, what is the average force of impact neglecting air resistance.

Here’s what I did.

d=\frac{v^2sin2\theta}{g}
200m=\frac{v^2sin90}{g}
1960=v^2
v=44.3m/s

I=mv-mv
I=0.046kg*44.3m/s
I=2.04Ns

I=Ft
2.04Ns =F(0.007s)
F=291.4N

Is this correct?
 
Last edited:
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I didn't check your arithmetic, but your method is correct.
 
You ususally use a pitching wedge to get a 45 degree angle. 45 degrees - 200 m? That is some wedge shot.

AM
 

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