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Godad
Oct9-04, 02:40 PM
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
Oct9-04, 03:06 PM
Use energy conservation to find

v = \sqrt {2 g h}

after the bounce.