1. It rebounds vertically to a height of 1.5m.
A 0.3kg ball is dropped from a height of 2m onto a hard surface. It rebounds vertically to a height of 1.5m. If it is in contact with the surface for 20ms, what is the average force exerted on it by the surface?
Break it down:
A 0.3kg ball
is dropped
from a height of 2m
onto a hard surface.
It rebounds vertically to a height of 1.5m.
If it is in contact with the surface for 20ms,
what is the average force exerted on it by the surface?
m = 0.3kg; mass of the ball
Free fall (initial speed is zero)
h1 = 2m; the height above the surface that ball starts moving downward. 
Hard surface means that surface will not change in shape or move when the ball is in contact with that. We will discuss it later in this chapter when we study collisions. 
h2 = 1.5m; the height above the surface that the ball travels upward.
Time interval; t = 20ms = 20*10-3 = 0.2sec 
Fave =?; we use impulse-momentum theorem to find average force. 
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2. Vi is initial speed just before the ball hits the surface.
A 0.3kg ball is dropped from a height of 2m onto a hard surface. It rebounds vertically to a height of 1.5m. If it is in contact with the surface for 20ms, what is the average force exerted on it by the surface?
Solution:
Draw a diagram
h1 = 2m
h2 = 1.5m Vf Vi
(1)
(2)
Vi is initial speed just before the ball hits the surface.
Vf is final speed immediately after the ball hits the surface and rebounds.
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3. Use the impulse-momentum theorem
A 0.3kg ball is dropped from a height of 2m onto a hard surface. It rebounds vertically to a height of 1.5m. If it is in contact with the surface for 20ms, what is the average force exerted on it by the surface?
Solution:
Use the impulse-momentum theorem
Impulse = ΔP; Impulse = F * Δt and ΔP = Pf - Pi
Pf = mVf; final momentum after the ball hits the surface.
Pi = mVi; initial momentum just before the ball hits the surface.
To calculate initial and final momentums we need to find initial and final speeds (velocity)
We use the principal of conservation of mechanical energy;
Before the ball strikes the surface;
EP = EK  [m * g * h] = (mVi2) /2  Vi = ( 2*g*h)1/2  Vi = (2* 9.8* 2)1/2
 Vi = 6.26m/s (this is the magnitude of the velocity. We know its direction is downward.)
Vi = m/s
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4. Calculate initial and final momentums
A 0.3kg ball is dropped from a height of 2m onto a hard surface. It rebounds vertically to a height of 1.5m. If it is in contact with the surface for 20ms, what is the average force exerted on it by the surface?
Solution:
After strike;
EK = EP  (mVf2) /2 = [m * g * h]  Vf = (2 * g * h)1/2  Vf = (2* 9.8* 1.5)1/2
 Vf = 5.24m/s or
Calculate initial and final momentums
m = 0.3kg P = mV Vi = m/s Vf = m/s
Pi = m*Vi  Pi = 0.3 * (-6.26)  Pi = -1.88kgm/s (downward)
and Pf = m*Vf  Pf = 0.3*(+5.42)  Pf = +1.63kgm/s (upward)
Vf = m/s (upward)
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5. Use the impulse-momentum theorem to find the impulse
A 0.3kg ball is dropped from a height of 2m onto a hard surface. It rebounds vertically to a height of 1.5m. If it is in contact with the surface for 20ms, what is the average force exerted on it by the surface?
Solution:
Use the impulse-momentum theorem to find the impulse
Impulse = ΔP = Pf - Pi  Impulse = (-1.88) 
Use impulse equation to find the average force
during the contact time interval which is Δt = 0.02sec
Impulse = Fave * Δt  Fave= Impulse/Δt
 Fave = 3.51/0.02
Impulse = 3.51kgm/s
The force is exerted on the ball by the surface during contact (collision)
 Fave= 175.5N
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