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?

Part a)

Part b)

Since it rebounds upwards so it's acceleration is in upwards direction

Explanation:

Velocity of the ball just before it will collide with the floor when it is dropped from height 4.00 m

So we will have

so we have

When ball rebound to height h = 2.00 m

so here we have

Part a)

For magnitude of average acceleration

Part b)

Since it rebounds upwards so it's acceleration is in upwards direction

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j