A steel ball is dropped onto a hard floor from a height of 1.95 m and rebounds to a height of 1.87 m (assume positive direction is upward)

A. Caculate it's velocity just before it strikes the floor
B. calculate it's velocity just after it leaves the floor on its way back up
C. Calculate it's acceleration during contact with the floor if that contact lasts .0800ms
D. How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid

To calculate the answers to these questions, we can use the principles of conservation of energy and the equations of motion.

A. To calculate the velocity just before the ball strikes the floor, we can use the equation of motion:

v^2 = u^2 + 2as

where:
- v is the final velocity (unknown)
- u is the initial velocity (unknown)
- a is the acceleration due to gravity (-9.8 m/s^2)
- s is the distance traveled (1.95 m)

Plugging in the known values, we get:

0 = u^2 + 2(-9.8)(1.95)

Simplifying the equation, we have:

u^2 = 19.08

Taking the square root, we find:

u ≈ 4.37 m/s (upward)

Therefore, the velocity of the ball just before it strikes the floor is approximately 4.37 m/s upwards.

B. To calculate the velocity just after it leaves the floor on its way back up, we can use the same equation of motion. However, this time the final velocity (v) is the negative of the initial velocity (u) since the direction is reversed:

v^2 = u^2 + 2as

0 = u^2 + 2(-9.8)(1.87)

Simplifying the equation, we have:

u^2 = 36.526

Taking the square root, we find:

u ≈ 6.05 m/s (downward)

Therefore, the velocity of the ball just after it leaves the floor on its way back up is approximately 6.05 m/s downward.

C. To calculate the acceleration during contact with the floor, we can use the equation of motion:

v = u + at

where:
- v is the final velocity (0 m/s since it comes to rest)
- u is the initial velocity (4.37 m/s)
- a is the acceleration (unknown)
- t is the time of contact (0.0800 ms or 0.0800s)

Plugging in the known values, we get:

0 = 4.37 + a(0.0800)

Simplifying the equation, we find:

a = -54.63 m/s^2

The negative sign indicates that the acceleration is in the opposite direction to the initial velocity. Therefore, the acceleration during contact with the floor is approximately -54.63 m/s^2.

D. To determine how much the ball compresses during its collision with the floor, we need to know the coefficient of restitution or the ratio of the relative velocities of separation and approach. However, assuming the floor is absolutely rigid, the collision is considered to be perfectly elastic, meaning there is no loss of kinetic energy. In this case, the ball does not compress at all.

Therefore, the ball does not compress during its collision with the floor, assuming the floor is absolutely rigid.