To test the quality of a tennis ball, you drop it onto a floor from a height of 4.00m . It rebounds to a height of 2.00 m. If the ball is in contact with the floor for 12.0ms,
(a) What is the magnitude of its average acceleration during the contact?
(b) is the acceleration up or down?

Question

anonymous · Answer

We can use relation: $$v ^{2} = v _{0}^{2} + 2ah$$
And I assume a = 9.8 m/s^2

1. Velocity before contact:
Initial velocity is zero anyway. So, final velocity before contact is: $$\sqrt{2ah}$$ 
Vi =  8.85 m/s

2. Velocity after contact:
Here, final velocity after rebound is zero - at the peak. So, the relation is same as above
Vf = -6.26 m/s (-ve since it is going up)

a. Average acceleration during contact:
   a = (vf - vi)/t = (-6.2 -8.8)/0.012 =  -1259.61 m/s^2

b. Clearly, the acceleration is up. This is shown by the -ve sign (since i took upwards as -ve). In addition, the ball's velocity was changed from downward to upward direction. This can happen when acc is upward.