Question

A tennis ball is dropped from 1.47 m above
the ground. It rebounds to a height of 1 m.
With what velocity does it hit the ground?
The acceleration of gravity is 9.8 m/s
2
. (Let
down be negative.)
Answer in units of m/s.
021 (part 2 of 3) 10.0 points
With what velocity does it leave the ground?
Answer in units of m/s.
022 (part 3 of 3) 10.0 points
If the tennis ball were in contact with the
ground for 0.00728 s, find the acceleration
given to the tennis ball by the ground.
Answer in units of m/s
2
.

Answer 1

V at ground = sqrt(2 g h)

g * Height attained = (1/2) v^2

change in momentum = m * sum of velocity down and velocity up
force = m a = change in momentum / change in time = m (|v|+ |V|) / .00728
so
a = the sum of the absolute values of velocity down and up / .00728

Answer 2

1. V^2 = Vo^2+2g*h = 0+19.6*1.47 = 28.8
V = 5.37 m/s.

2. V^2 = Vo^2+2g*h
0 = Vo^2+(-19.6)1
Vo = 4.43 m/s.

3. V = Vo+a*T
4.43 = 5.37+a*0.00728
a =