posted by Lindsay on .
A golf ball released from a height of 1.80 m above a concrete floor, bounces back to a height of 1.06 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?
Do I need to know the velocity in order to solve this? I'm completely lost.
Determine the velocity just as the ball makes first contact with the floor. Determine the velocity as the ball just just leaves the floor. Calculate the delta velocity. Divide this by the delta t to get (delta v)/((delta t) which is average acceleration. Look for the forumula relating velocity to acceleration and distance. It will look something like v=sqrt(2*?*?)
What information must I use to get the velocities?
Velocity= sqrt(2 * acceleration * distance). This is a special case where the velocity is zero at one end of the path which is the case here.
acceleration is gravity
the two distances are given. The signs of the two velocities are different. Remember that when you subtract them.
Then divide the difference by the delta time to get a m/(s^2) acceleration.