Answer:
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.
Answer:
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Explanation:
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Answer:
Explanation:
a charged and uncharged object attratct eachother that is the answer your welcome
Here are the steps you would need to follow:
#1). Define what 'the 'X is, and how it's related to the ball.
#2). Be clear on how 'the X' is related to the 'known velocity'.
#3). Identify how the 'known velocity' is related to the action of the ball when it's launched.
With this information in front of you, you'll have a much better chance
of answering the question.
With none of it in front of me, I have no chance at all.
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