Solving Impact Acceleration Homework

[Argento7]

Homework Statement

A golf ball released from a height of 2.02 m above a concrete floor, bounces back to a height of 0.86 m. If the ball is in contact with the floor for 3.57 ms( or .00357s), what is the magnitude of the average acceleration a of the ball while it is in contact with the floor?

Homework Equations

Is the .86m bouncing back up irrelevant?

The Attempt at a Solution

I found out that the initial position is 2.02m and the initial position is 0m and I cut the time in half and used 9.8 as my acceleration. I'm just really stuck and have gotten myself caught up in so many different equations my brain can not handle it anymore. Any help would work please.

 

[LowlyPion]

Homework Statement

A golf ball released from a height of 2.02 m above a concrete floor, bounces back to a height of 0.86 m. If the ball is in contact with the floor for 3.57 ms( or .00357s), what is the magnitude of the average acceleration a of the ball while it is in contact with the floor?

Homework Equations

Is the .86m bouncing back up irrelevant?

The Attempt at a Solution

I found out that the initial position is 2.02m and the initial position is 0m and I cut the time in half and used 9.8 as my acceleration. I'm just really stuck and have gotten myself caught up in so many different equations my brain can not handle it anymore. Any help would work please.

You need how far it bounces back to determine the speed it left the floor on the rebound, just as you need the height to determine the speed it hit the floor.

 

[baba89]

As a scientist, it is important to approach problems systematically and use the appropriate equations and methods to find a solution. In this case, we can use the equation for average acceleration, which is a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity, and t is the time. In this problem, the initial velocity is 0 m/s since the ball is released from rest, and the final velocity is unknown. Therefore, we need to find the final velocity first.

To find the final velocity, we can use the equation vf^2 = vi^2 + 2ad, where d is the displacement, which in this case is the height of the ball after bouncing, 0.86 m. Plugging in the values, we get vf = √(0^2 + 2(9.8)(0.86)) = 3.17 m/s.

Now that we have the final velocity, we can plug it into the equation for average acceleration. We also need to convert the time from milliseconds to seconds, so it becomes 0.00357 s. Therefore, the average acceleration is a = (3.17 - 0)/0.00357 = 887.3 m/s^2.

In conclusion, the magnitude of the average acceleration of the ball while it is in contact with the floor is 887.3 m/s^2. It is important to note that this is the average acceleration, which means that the ball may have experienced different accelerations during its contact with the floor, but this is the overall magnitude.