To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. it rebounds to a height of 2.00 m. if the ball is in contact with the floor for i2.0 ms, (a) what is the magnitude of its average acceleration during that contact and (b) is the average acceleration up or down?

Part a)

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

Part a)

Average acceleration is given as

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

current flows in a wire

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day