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Average acceleration

A ball is dropped from 4 ft and rebounds back to 3 feet. If the ball was in contact with the ground for 0.01 seconds, then what is the average acceleration of the ball during contact? The answer on the back of the book is 3000 ft/sec^2. So average acceleration is defined as [tex] \frac{\Delta V}{\Delta t} [/tex]. So [tex] \frac{x}{0.01} [/tex]. Then [tex] \Delta V = 30 [/tex]. How would you get 30?

Thanks
 

OlderDan

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You need to calculate the speed at impact for a ball dropped from 4 feet, and the speed it would need to have when leaving the floor to rise to a height of 3 feet. Take the difference between the two velocities (speed and direction) to get [tex] \Delta V [/tex]
 

HallsofIvy

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A ball falls a distance of 4 feet. You know d= -16t2+ 4 and d= 0 when the ball hits the floor. Solve -16t2+ 3= 0 to find how long the ball was falling. Of course, V= -32t so you can find the speed of the ball as it hits the floor.
(And it will be negative, of course.)

On the way up, V= -32t+ V0 and d= -16t2+ V0t where V0 is the speed just as the ball is leaving the floor. Of course, at the ball's highest point V= 0 so you have the two equations -32t+ V0= 0 and -16t2+ V0t= 3 to solve for V0. (Since there are two equations you can solve for both V0 and t but V0 is all you need to find.)

The change in the ball's velocity during the bounce is V0 minus the first V you found (which, remember, was negative).
 

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