# Golf ball

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.

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

learningphysics
Homework Helper
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

learningphysics
Homework Helper
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

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

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?

learningphysics
Homework Helper
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.