# Golf ball

1. Aug 30, 2007

### chaotixmonjuish

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.

2. Aug 31, 2007

### learningphysics

What is the velocity before the ball hits the ground when it's dropped from 1.76m?

3. Aug 31, 2007

### learningphysics

You should be familiar with this equation:

$$v_{2}^2 = v_{1}^2 + 2as$$ where s is the displacement.

4. Aug 31, 2007

### chaotixmonjuish

I actually gave you all my givens, unless I'm suppose to calculate velocity.

5. Aug 31, 2007

### chaotixmonjuish

Is it 9.799 m/s^2

6. Aug 31, 2007

### chaotixmonjuish

okay, i keep getting ansewrs in the 23.86 range, however they seem to be wrong

7. Aug 31, 2007

### learningphysics

Yes, calculate the velocity right before it hits the ground... use the formula I posted.

8. Aug 31, 2007

### chaotixmonjuish

okay using that formula, i got 17.924 as the average acceleration

9. Aug 31, 2007

### chaotixmonjuish

actually i took the acceleration of both sides and then took the average

10. Aug 31, 2007

### learningphysics

No, that's not right. Can you post your work? What did you get for the velocity before the ball hits the ground?

11. Aug 31, 2007

### chaotixmonjuish

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. Aug 31, 2007

### learningphysics

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