Dropped Tennis Ball Homework: Speed & Force Calculation

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A 0.4 kg tennis ball dropped from a height of 6.2 m reaches a speed of approximately 11 m/s upon contact with the floor, calculated using the equation v = (2gx)^(0.5). The discussion then shifts to calculating the force exerted by the floor on the ball, considering the ball compresses 0.6 cm upon impact. An initial attempt at calculating the acceleration during contact yields 9.76 m/s², leading to a force of 3.9 N, but this is deemed incorrect. Suggestions are made to use energy relations and consider momentum for a more accurate approach. The conversation emphasizes the importance of verifying calculations and understanding the physics involved.
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Homework Statement


A 0.4 kg tennis ball is dropped from rest at a height of 6.2 m onto a hard floor.


Homework Equations


a) What is the speed of the ball at the instant of contact with the floor?
v=(2gx)^.5
v=(2*9.81*6.2m)^.5
=11 m/s
b) A flash photograph shows that the ball is compressed a maximum of 0.6 cm when it strikes the floor.
Assuming that the acceleration of the ball is constant during its contact with the floor, what force does the floor exert on the ball?

The Attempt at a Solution


So for this question, i used the velocity as a function of position equation: V^2=Vo^2+2a(Delta)y
Then i solved for a a=v^2-Vo^2/2(Delta)y
a=11^2-0/2(6.194)=9.76 m/s^2
F=ma=.4kg(9.76 m/s^2)=3.9 N
But this answer is incorrect. I think i have the right idea, maybe i am using the wrong equation or did i make a mistake?
 
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For part a) what is the easiest way is energy relations. You got the equations on this part. Double check your numbers, though. I haven't a calculator, but it's a good idea.

For part b), there is a better relation. Ask yourself what the center of mass is doing on impact. And then think about momentum.
 
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