Calculating Average Acceleration of a Golf Ball

AI Thread Summary
To calculate the average acceleration of a golf ball bouncing from a height of 1.76 m to 0.80 m, the initial velocity before impact is determined using the formula v² = u² + 2as, yielding approximately 5.873 m/s. The ball is in contact with the floor for 4.62 ms, which is crucial for calculating average acceleration. The average acceleration can be computed by analyzing the change in velocity during the contact time. The discussion emphasizes the need to calculate both the impact velocity and the velocity after the bounce to fully understand the motion. Understanding these concepts is essential for accurately determining the average acceleration of the ball.
chaotixmonjuish
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A golf ball released from a height of 1.76 m above a concrete floor, bounces back to a height of 0.80 m. If the ball is in contact with the floor for 4.62 ms, what is the magnitude of the average acceleration a of the ball while it is in contact with the floor?


4.9x^2-1.76

I used this to determine how long it would take for the ball to get to .08m. Apparently it will take .7228 s.

Outside of that, I'm not really sure what to do. Is there a way to relate position to velocity? And what exactly is the magnitude of the average acceleration? I haven't heard that term at all until this question.
 
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What is the velocity before the ball hits the ground when it's dropped from 1.76m?
 
You should be familiar with this equation:

v_{2}^2 = v_{1}^2 + 2as where s is the displacement.
 
I actually gave you all my givens, unless I'm suppose to calculate velocity.
 
Is it 9.799 m/s^2
 
okay, i keep getting ansewrs in the 23.86 range, however they seem to be wrong
 
chaotixmonjuish said:
I actually gave you all my givens, unless I'm suppose to calculate velocity.

Yes, calculate the velocity right before it hits the ground... use the formula I posted.
 
okay using that formula, i got 17.924 as the average acceleration
 
actually i took the acceleration of both sides and then took the average
 
  • #10
No, that's not right. Can you post your work? What did you get for the velocity before the ball hits the ground?
 
  • #11
4.9x^2-1.76=0
x=.5993

9.8(.5593)=5.873 m/s at the point of impact

would i have to do something similar for the other side?
 
  • #12
chaotixmonjuish said:
4.9x^2-1.76=0
x=.5993

9.8(.5593)=5.873 m/s at the point of impact

would i have to do something similar for the other side?

Exactly... find the initial velocity during the path when the ball goes upwards to 0.80m. ie: the velocity right after it bounces up.
 
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