Tennis Ball Drop II: Find Force Exerted on Floor

[PhysicslyDSBL]

Homework Statement

A 0.5 kg tennis ball is dropped from rest at a height of 5.6 m onto a hard floor.

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?

(This is a problem in one-dimensional kinematics with constant acceleration. What is the acceleration of the ball when it is in contact with the floor?)

f = N

Homework Equations

In the first part of the problem, I got velovity, 10.48m/s. I am now looking for acceleration, a. I attempted to find a using:

a = (v^2 - u^2)/2s

v = final velocity, 10.48
u = initial velocity, 0
s = displacement, 5.6

The answer is a = 9.80 and multiplying by mass 0.006 (0.6cm converted to meters) to get 0.059 as F.

The Attempt at a Solution

Please help?!

 

[Prologue]

If the acceleration is constant when it hits the floor then the ball has changed from 10.48 m/s to 0 m/s within a distance of 0.6 cm.

You need to know two things here:

You want to find out the time that it took for the ball to go from 10.48 m/s down to 0, but you only have a distance that relates them. How do you change a distance into time when a velocity and distance is involved?

distance = rate * time so --> time = distance/rate

You have the distance but you don't have the rate, how can you get the rate? (Hint think averages.)


Once you have that then use the definition of average acceleration to give you the answer.

change in velocity/change in time = constant acceleration

 

[PhysicslyDSBL]

So the average speed at which the ball is moiving is the average rate. To calculate this I need to divide the distance by time.

My attempt at finding time is:

t = sqrt 2s/a

s = 5.6
a = 9.81

the answer is t = 1.14

I then take this quantitiy to use:

d/t = r

d = 5.6
t = 1.14

the answer is 4.91

I'm pretty sure I am still missing something.