Calculating Average Acceleration of a Golf Ball

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Homework Help Overview

The discussion revolves around calculating the average acceleration of a golf ball that is dropped from a height of 1.76 m and bounces back to a height of 0.80 m. Participants are exploring the relationships between height, velocity, and acceleration during the ball's contact with the floor.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to calculate the velocity of the ball just before impact and are discussing the use of kinematic equations. Questions are raised about the meaning of average acceleration and how to relate position to velocity.

Discussion Status

There is an ongoing exploration of different calculations and approaches to determine the average acceleration. Some participants are sharing their calculations and questioning the validity of their results, while others are suggesting the use of specific equations to find necessary values.

Contextual Notes

Participants mention specific time constraints and the need to calculate velocities at different points in the ball's trajectory. There is an indication of confusion regarding the terms and concepts involved in the problem.

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:

[tex]v_{2}^2 = v_{1}^2 + 2as[/tex] 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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