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

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2