Dropped Tennis Ball Homework: Speed & Force Calculation

[mrnastytime]

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?

 

[Mr.Amin]

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.

 

[thomate1]

Your approach and equations are correct. However, there may be a mistake in your calculation. When solving for acceleration, you should use the change in position (Delta y) as 0.006 m, since the ball is compressed by 0.6 cm (0.006 m) and not 6.194 m. This would give an acceleration of 97.6 m/s^2 and a force of 39.04 N. Also, make sure to use the correct units in your calculations to avoid any errors. Keep up the good work!