How high does the ball rebound?

AI Thread Summary
The discussion focuses on calculating the rebound height of a rubber ball after it is dropped from a height of 2.0 m. The user has already solved the first six parts of the homework but is seeking assistance specifically for part g, which asks how high the ball rebounds. Key equations related to momentum and velocity are provided, but the user is struggling with the calculation of rebound height. Responses emphasize that the graph of the normal force is only relevant for impulse calculations, while the rebound speed can be derived from the momentum change. The user is encouraged to determine the actual speed of the rebounding ball to find the correct rebound height.
edr2004
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I ONLY NEED HELP WITH LETTER G. I ALREADY HAVE THE CORRECT ANSWERS FOR A-F.
I posted all of the questions so you guys can see all the information I have.

Homework Statement



A rubber ball with a mass of 200 g is released from rest from a height of 2.0 m. It falls to the floor, bounces, and rebounds. The graph at right depicts the magnitude of the upward normal force that the floor exerts on the ball at various instants in time. The graph only shows the narrow window of time surrounding the interval when the ball was in contact with the floor.

a.) How long does it take the ball to fall the 2.0 m to the floor?
b.) How fast is the ball traveling just before it hits the floor?
c.) What is its momentum just before hitting the floor (using a coordinate system in which the positive-y axis points in the upward direction)?
d.) What is the impulse on the ball by the floor during the 10 ms the ball is in contact with the floor?
e.) What is the impulse on the ball by the earth’s gravitational pull during the same 10 ms?
f.) By how much does the ball’s momentum change as a result of this 10-ms period?
g.) How high does the ball rebound?


Homework Equations


Change of momentum = mass * change of velocity
V2^2 = 2a*change of position


The Attempt at a Solution


Change of velocity = (2.4kg.m/s) / (0.2kg) = 12m/s
(12m/s)^2 = 2 (9.81m/s2) * change in position
CHange in position = (144m2/s2) / 2*9.81m/s^2 = 7.34m , which is not correct.
Also I was able to calculate Acceleration from the impulse on ball by the floor. I am not sure if and/or how I would use that acceleration.
Anyway I would appreciate any help.
 
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I do not see your graph.
 
I attached the graph (it will open as a Word document). For some reason I can't paste it with my message.
 

Attachments

First things first. Answer a, b, & c: those question have nothing to do with your graph; treat the ball as any other falling body. The graph is only relevant for answering d. (For that you'll need the definition of impulse.)
 
Thanks for replying but again I ONLY NEED HELP WITH G, which asks how high the ball will rebound. I already have the correct answers for a-f. I only posted the other questions so you guys would know what information I was given. So if you could look under my orignal post at the attempt to solve the problem and offer some suggestions I would appreiciate it.
THANKS
 
Next time BE CLEARER. :smile:

edr2004 said:

The Attempt at a Solution


Change of velocity = (2.4kg.m/s) / (0.2kg) = 12m/s
(12m/s)^2 = 2 (9.81m/s2) * change in position
CHange in position = (144m2/s2) / 2*9.81m/s^2 = 7.34m , which is not correct.
Find the actual speed of the rebounding ball, not its change in speed. (You can use the results of f to find that rebound speed.)
 
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