Consider a ball of 0.22 kg, initially at rest, is dropped from an initial height of 1.80 m. It rebounds back after colliding with the floor to a final height of 1.50 m. Determine the impulse on the ball delivered by the floor.

The impulse on the ball delivered by the floor is 2.52 kg-m/s.

Explanation:

Given that,

Mass of the ball, m = 0.22 kg

It  is dropped from an initial height of 1.80 m. It rebounds back after colliding with the floor to a final height of 1.50 m. Initial velocity and final velocity can be calculated using conservation of energy as :

Final velocity,

As the ball rebounds, v = -5.47 m/s

We need to find the impulse on the ball delivered by the floor. We know that impulse is equal to the change in momentum as follows :

So, the impulse on the ball delivered by the floor is 2.52 kg-m/s.

ok

explanation:

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.