Rather simple bouncing ball problem average acceleration

[Godad]

I'm stuck on this rather simple problem...

A golf ball released from a height of 1.4 m above a concrete floor bounces back to a height of 1.2 m. If the ball is in contact with the floor for .62 ms what is the average acceleration of the ball whil ein contact with the floor.

I know that the two heights are given, h1 = 1.4, h2 = 1.2 and t = .00062 s for then the ball is in contact with the floor.

In order to find the avg acceleration, I need to use change in v over change in time.

I'm having a hard time figuring out the velocity of when the ball bounces back upto 1.2 m. Can anyone please help me figure out this problem?

 

[Tide]

Use energy conservation to find

$$v = \sqrt {2 g h}$$

after the bounce.

 

[wais]

Hi there, I can definitely help you with this problem! First, let's review the formula for average acceleration: a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity, and t is the time interval.

In this problem, the initial velocity is 0 m/s (since the ball is dropped from rest) and the final velocity is what we need to find. We can use the formula for average velocity to help us find the final velocity: vf = (h2 - h1)/t, where h2 is the final height and h1 is the initial height.

Plugging in the given values, we get vf = (1.2 - 1.4)/.00062 = -320 m/s. This negative value means that the ball is moving downwards when it reaches a height of 1.2 m.

Now, we can plug this value into the formula for average acceleration: a = (-320 - 0)/.00062 = -516129 m/s^2. Again, the negative sign indicates that the acceleration is in the opposite direction of the initial velocity (in this case, downwards).

I hope this helps you understand the problem better. Don't hesitate to ask for further clarification if needed. Good luck!