# Dropped Tennis Ball question

Can someone please check my equations to see if I am going about this the right way.

A 0.4 kg tennis ball is dropped from rest at a height of 3.9 m onto a hard floor.

a) What is the speed of the ball at the instant of contact with the floor? I solved to be 8.743

A flash photograph shows that the ball is compressed a maximum of 0.6 cm when it strikes the floor.

b) Assuming that the acceleration of the ball is constant during its contact with the floor, what force does the floor exert on the ball?

I used my v final from part a as my v initial for part b. So
v initial = 8.743
v final = 0
delta x = 3.9

using the formula vf2=vI2+2axdelta x
which gave me 0=(8.734)2+(2)ax(.6) = 63.7

then using F=ma+9.8m
gave f= (0.4)(63.7)+(9.8)(0.4) so F=29.4

but this is incorrect. Can anyone point out what I'm missing?

## Answers and Replies

Assuming you used the right numbers, it looks like number 1 is probably correct.

For part two the tennis ball is compressed, that's interesting, it's almost spring-like...

The only way I now how to solve force of a spring is using Hookes Law, and I cant seem to make that work either.

Is there any other advice you can give me?
Could it have something to do with my original displacement is in m and the second part is in cm?

anyone?

alphysicist
Homework Helper
Hi jspek9,

The only way I now how to solve force of a spring is using Hookes Law, and I cant seem to make that work either.

Is there any other advice you can give me?
Could it have something to do with my original displacement is in m and the second part is in cm?

That's right; you have a problem with the units in your work. When you calculated this line in your original post:

using the formula vf2=vI2+2axdelta x
which gave me 0=(8.734)2+(2)ax(.6) = 63.7

the 8.734 is in m/s, and the .6 is in cm, so the 63.7 will not be in m/s$^2$. You probably want to convert the 0.6cm to meters.