Calculating Impulse of a Ball Dropped from Rest

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To calculate the impulse given to a 0.150 kg ball dropped from a height of 1.25m, the relevant equations include impulse (I = Δp) and momentum (p = mv). The ball rebounds to a height of 0.960m, and the key to solving the problem is determining the velocities just before and after the collision with the floor. Kinematic equations can be used to find these velocities, specifically vf² = vi² + 2a(xf - xi), without needing to know the time. This approach will allow for the calculation of Δv and subsequently Δp, leading to the impulse value.
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Homework Statement


A 0.150 kg ball is dropped 1.25m from rest. It rebounds to a height of 0.960m. What is the impulse given to the ball by the floor?


Homework Equations


I=\Deltap and p=mv were the equations I was trying to use. I was pretty sure I should use F=ma since the only other thing that I know is acceleration due to gravity. I don't know anything about time so I ruled out \int\sumFdt.


The Attempt at a Solution

 
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mickellowery said:
I=\Deltap and p=mv were the equations I was trying to use.
Good. So how can you figure out the velocity just before and just after it hits the floor?

Hint: When the ball is in the air, it's just a projectile.
 
Using I = Δp is the correct approach. Can you find what v just before and just after the collision? If so, then you can find Δv and from this Δp.
 
So can I use the kinematics equations? Since I don't know anything about time can I use vf2= vi2+2a (xf-xi) to find the velocity?
 
mickellowery said:
So can I use the kinematics equations? Since I don't know anything about time can I use vf2= vi2+2a (xf-xi) to find the velocity?
Sounds good to me.
 
thanks kindly to all!
 
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