# Homework Help: 2D Motion Physics Problem

1. Nov 17, 2008

### chrozer

1. The problem statement, all variables and given/known data
An arrow is fired with a horizontal spped of 89 m/s directly at a target 60 m away. When it is fired, the arrow is 1 m above the ground. How far short of the target is it when it strikes the ground?

2. Relevant equations
D = vt
D = Vit + (1/2)at2
Vf = Vi + at

3. The attempt at a solution
Basically my teacher says to set up a table, one side horizontal values and the other vertical values.

The values I know for:
• Horizontal - velocity = 89, distance = 60
• Vertical - acceleration = -9.8, distance = 1

Then that is where I get stuck...help? BTw the anser is 20 m short, but my teacher wants me to show work.

Last edited: Nov 17, 2008
2. Nov 17, 2008

### Redbelly98

Staff Emeritus
There are some more items to fill out in your table.

Horizontal: time?
Vertical: time?

Also, there are a couple of problems with the values you do have.

1. If the arrow falls short of the target, it does not go the full 60m distance.

2. Acceleration is -9.8 m/s^2, so you are using upward=positive, downward=negative. But then you say the vertical distance travelled is +1m upward. Gotta watch those +/- signs.