- #1

Yousufshad

- 28

- 0

## Homework Statement

A 178.0g ball is dropped from a height of 2.99m, bounces on a hard floor, and rebounds to a height of 1.36m. The impulse received from the floor is shown below.

(Is a picture of a graph, Y-axis is force, X-axis is time. No numbers are labelled and is simply a line starting at 0 and goes to a peak and back down to zero like a triangle)

What maximum force does the floor exert on the ball if it is exerted for 2.00ms

## Homework Equations

Impulse = Change in Momentum

Momentum = mv

Ui = mgy

KE = 1/2 mv^2

## The Attempt at a Solution

[/B]

Ui = mgy

=5.2157

1/2 mv^2 = 5.2157

v=7.655m/s

(to find velocity just as ball hits the ground)

Pi = (0.178)(7.655)

=1.3626

(Initial Momentum)

Finding velocity just as it bounces back from floor

Uf = KEi

v=5.1629m/s

Pf = 0.919

(final momentum)

Change in momentum = 0.433

DeltaP(change in momentum) = 1/2 Fmax (0.002)

Fmax = 443N (not correct answer) where did I go wrong?