# Homework Help: Relative motion problem

1. Dec 28, 2012

### ShizukaSm

1. The problem statement, all variables and given/known data
To an observer not in motion in the ground, the trajectory of a projectile is described by:
$y = x - 10^{-3}x^2$(In SI units)
Where it was adopted x = 0 and y = 0 for the origin point. Determine the trajectory equation for an observer that moves at 20sqrt(2) m/s in relation to the ground, in the positive x axis.

(Answer is $y = \frac{5}{3}x - \frac{1}{360}x^2$)

2. Relevant equations

Only the equations already given.

3. The attempt at a solution

I have no idea whatsoever on how to deal with this problem. I have learned how to deal with relative position problems, and relative velocity problems, but I have no idea how to change the equation themselves and make them reflect that.

Last edited: Dec 28, 2012
2. Dec 28, 2012

### hgfhh123

Look at each motion separately. First of all, what is the x component of the velocity relative to the stationary observer, from your equation? Then, if the observer is moving with 20√2 m/s, what is the x component relative to him? Do the same for the y component.

Think about what that graph is actually telling you about these values.

3. Dec 29, 2012

### ShizukaSm

I apologize, but I do not understand what you mean. Could you maybe rephrase it, in case I am missing something obvious?

I tried taking the inverse function (So I would have the movement in function of x), but still, that didn't help me.

4. Dec 29, 2012

### rbrayana123

Hmm.. if you start with your basic kinematics relationships, you should be able to construct a relationship between y and x involving the initial velocity angle, initial velocity magnitude and the constant acceleration.

However, in this case, you don't know the initial angle, velocity and accelerations. So, it's a matter of working it out the other way.

EDIT: Also you'll probably be working under the assumption that there is only a deceleration of g in the y direction. With all that, you should get the raw materials for which you can answer the question.

Last edited: Dec 29, 2012