# Impulse of a falling ball under gravity onto a solid surface

## Homework Statement

A 0.5-kg rubber ball is dropped from rest a height of 19.6 m above the surface of the Earth. It strikes the sidewalk below
and rebounds up to a maximum height of 4.9 m. What is the magnitude of the impulse due to the collision with the sidewalk?

## Homework Equations

J=$$\Delta$$P

## The Attempt at a Solution

mgh=.5mv2
v=$$\sqrt{}2gh$$

Just before collision:
v=$$\sqrt{}2*9.8*19.6$$
v=19.6 m/s

Just after collision:
v=$$\sqrt{}2*9.8*4.9$$
v=9.8 m/s

J=m(vafter-vbefore)=.5(9.8-19.6)=4.9 kg m/s

The answer is 14.7 N.s, so I'm not sure what I'm missing.

Related Introductory Physics Homework Help News on Phys.org

Well velocity is a vector - is this how you would subtract them?

J=.5[9.8-(-19.6)]=14.7 kg m/s

Wow that was a stupid mistake, thanks!