A steel ball is dropped onto a hard floor from a height of 2.15 m and rebounds to a height of 2.09 m. (Assume that the positive direction is upward.)

(a) Calculate its velocity just before it strikes the floor.
-6.49 m/s
(b) Calculate its velocity just after it leaves the floor on its way back up.
m/s
(c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms.
m/s2
(d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

b is a little tricky.

Notice the height an object gets depends on v^2 (1/2 mv^2=mgh)

so the velocity just after rebound, is propotional the the sqrtrt of the rebound(hf/hi)

b. v=+6.49*sqrt(2.09/2/15) (I did not check the answer in a

c. Force*time=changemomentum
mass*a*time=mass(answerinB-answerinA) solve for a.
notice you will ADD the two answers, the signs on the second term come from -(-6.49)
d I am not certain of what the instructor has in mind here.

The compression is based on the displacement of the ball during its time interacting with the floor. The initial velocity is the velocity of the ball just before it hits the floor, and the final velocity is 0. The acceleration is the acceleration while the ball is interacting with floor > (answer b- answer a)/ time on the floor

To solve this problem, we can make use of the equations of motion.

(a) To calculate the velocity just before the ball strikes the floor, we can use the equation for free fall motion:

v^2 = u^2 + 2as

where:
v = final velocity (unknown)
u = initial velocity (0 m/s, since the ball was dropped)
a = acceleration due to gravity (9.8 m/s^2)
s = distance traveled (2.15 m)

Rearranging the equation, we get:

v^2 = 0 + 2 * 9.8 * 2.15
v^2 = 42.49

Taking the square root of both sides, we find:

v ≈ -6.49 m/s

Therefore, the velocity just before the ball strikes the floor is approximately -6.49 m/s.

(b) To calculate the velocity just after the ball leaves the floor, we use the equation for free fall motion again:

v_f^2 = v_i^2 + 2as

where:
v_f = final velocity (unknown)
v_i = initial velocity (-6.49 m/s, since the ball rebounds in the opposite direction)
a = acceleration due to gravity (9.8 m/s^2)
s = distance traveled (2.09 m)

Rearranging the equation, we get:

v_f^2 = (-6.49)^2 + 2 * 9.8 * 2.09
v_f^2 = 42.2401 + 40.784
v_f^2 = 83.0241

Taking the square root of both sides, we find:

v_f ≈ 9.11 m/s

Therefore, the velocity just after the ball leaves the floor is approximately 9.11 m/s.

(c) The acceleration during contact with the floor can be calculated using the equation:

a = (v_f - v_i) / t

where:
a = acceleration (unknown)
v_f = final velocity (9.11 m/s)
v_i = initial velocity (-6.49 m/s)
t = time (0.0800 ms = 0.0800 * 10^-3 s)

Substituting the values into the equation, we get:

a = (9.11 - (-6.49)) / 0.0800 * 10^-3
a = 15.60 / 0.0800 * 10^-3
a ≈ 195,000 m/s^2

Therefore, the acceleration during contact with the floor is approximately 195,000 m/s^2.

(d) To calculate how much the ball compressed during the collision with the floor, assuming the floor is absolutely rigid, we can use the equation for elastic potential energy:

Elastic Potential Energy = 1/2 k x^2

where:
k = spring constant (unknown)
x = displacement/compression (unknown)

Since the floor is absolutely rigid, k is infinite, which means the ball doesn't compress at all. Therefore, the ball doesn't compress during its collision with the floor.

To answer these questions, we can use the principles of physics, specifically the laws of motion and energy conservation.

(a) To calculate the velocity just before the ball strikes the floor, we can use the concept of conservation of mechanical energy. The mechanical energy of the ball is conserved, neglecting any energy losses due to air resistance or friction. Therefore, the potential energy of the ball at the initial height must be equal to the sum of its kinetic energy just before it hits the floor.

The potential energy at the initial height is given by:
Potential energy = mass * gravity * height

Assuming the mass of the ball is known, and gravity is approximately 9.8 m/s^2, we can calculate the potential energy.

Next, we equate the potential energy to the kinetic energy just before impact, given by:
Kinetic energy = 0.5 * mass * velocity^2

We can rearrange this equation to find the velocity just before impact:
velocity = sqrt(2 * (potential energy / mass))

Plugging in the values gives us the answer.

(b) To calculate the velocity just after the ball leaves the floor on its way back up, we can again use the concept of conservation of mechanical energy. This time, we need to consider the potential energy at the rebound height and the kinetic energy at departure.

Using the same equation as in part (a), we can find the velocity just after the ball leaves the floor.

(c) To calculate the acceleration during contact with the floor, we need to use a different formula. We know that acceleration is the change in velocity over time. In this case, since the contact time is given, we can divide the change in velocity by the contact time to get the acceleration.

Acceleration = (change in velocity) / (contact time)

Substituting the given values will give us the result.

(d) To calculate the compression of the ball during its collision with the floor, we need to know the elastic properties of the ball. The compression is a measure of how much the ball gets deformed during the collision. Since the floor is stated to be absolutely rigid, it means that the ball does not compress at all.

Therefore, the compression would be zero in this case.

By following these explanations and using the relevant equations, you should be able to solve the given problem.