A tennis ball is dropped on to the floor from a height of 10 m. It rebounds to a
height of 2.5 m. If the ball is in contact with the floor for 0.01 seconds, what is
the average acceleration during the contact?
You have initial PE, so solve for velocity at floor. Then, you know PE at 2.5m, so you know initial rebound velocity.
acceleration= (Vrebound-Vinitial)/time remembering that Vrebound is exactly the opposite direction of Vinitialcontact....so they add.
To find the average acceleration during the contact, we can use the formula:
Average acceleration = (Change in velocity) / (Time taken)
To use this formula, we need to determine the change in velocity and the time taken.
First, let's calculate the change in velocity:
The ball is dropped from a height of 10 m and rebounds to a height of 2.5 m. The change in height is:
Change in height = Final height - Initial height
= 2.5 m - 10 m
= -7.5 m (negative because the ball is going upwards after the rebound)
Next, let's calculate the change in velocity using the formula:
Change in velocity = (Initial velocity) - (Final velocity)
Since the ball is dropped, the initial velocity is 0 m/s (at rest). The final velocity can be calculated using the equation:
Final velocity = (Change in height) / (Time taken)
= -7.5 m / 0.01 s
= -750 m/s (negative because the ball is going upwards after the rebound)
Now, we can find the average acceleration by dividing the change in velocity by the time taken:
Average acceleration = (Change in velocity) / (Time taken)
= (-750 m/s) / (0.01 s)
= -75,000 m/s² (negative because the ball is slowing down during the upward flight)
So, the average acceleration during the ball's contact with the floor is -75,000 m/s².