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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?