Average force between an object and non-movable object?

AI Thread Summary
To calculate the average force on a golf ball dropped from a height of 1.41 m that bounces back to the same height, the change in velocity at ground level must be determined. The initial velocity (Vi) is 0 m/s, and the final velocity (Vf) can be calculated using the equation Vf = √(2gh), where g is the acceleration due to gravity (9.8 m/s²). The impulse-momentum theorem states that impulse equals the change in momentum, which is given by Ft = m(Vf - Vi). The average force can then be calculated by rearranging the impulse equation. Understanding the velocities at ground level is crucial for solving the problem accurately.
sirfinklstin
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Homework Statement



A golf ball of mass 41.8 g is dropped from a height h = 1.41 m onto a hard floor and bounces back to exactly that same height. What is the average force (in N) on the ball during the 0.012 s that it was in contact with the floor?

Homework Equations



Impulse = Ft
Impulse = m(Vf - Vi)

The Attempt at a Solution


I am not quite sure how to proceed with this problem, becuase I am completely clueless as to how I should go about calculating the change in velocity, becuase if the ball returns to the same height, (no friction of air resistance) then arent the Vi and Vf both = 0?
 
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Hi sirfinklstin! :smile:
sirfinklstin said:
A golf ball of mass 41.8 g is dropped from a height h = 1.41 m onto a hard floor and bounces back to exactly that same height. What is the average force (in N) on the ball during the 0.012 s that it was in contact with the floor?

how I should go about calculating the change in velocity, becuase if the ball returns to the same height, (no friction of air resistance) then arent the Vi and Vf both = 0?

Yes, but you want the two velocities at ground level. :wink:
 
How would the velocities be calculated? 0 + (9.8 m/s)(1.41m)?
 
sirfinklstin said:
How would the velocities be calculated? 0 + (9.8 m/s)(1.41m)?

what? v = as ?? :confused:
 
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