# Shoot a Rocket off a Cart and through the Hoop

1. Sep 28, 2008

### Phoenixtears

1. The problem statement, all variables and given/known data
A 380 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exerts an 7.5 N thrust on the rocket. Your goal is to have the rocket pass through a small horizontal hoop that is 20 m above the launch point. At what horizontal distance left of the loop should you launch?
____m

2. Relevant equations

Rolling force= (coeffictient of rolling force(Mu))*(N)
F=Ma
Delta-x= V0*t + .5a(t^2)
Vf^2= V0^2 + 2ax
Vf= V0 + at

3. The attempt at a solution

I don't know how to attack this problem, but this is how I started. I first drew a force diagram that involves normal, weight, and rolling friction (involving the cart). But then I assumed that a completely separate force must be used for the rocket, because I'm assuming that the rocket is not strapped to the cart. So, that would involve normal, weight, thrust, and drag. Then I found the Mu first (even though it is technically known, I've seen problems that use weird coeffictients). In other words, 7.5= (9.8*380)*(Mu). Mu= .002- this involves the cart. I drew a table for delta-x, inital V, final V, acceleration, and time. I know that I solving for horizontal x and that I have verticle x (20). But what else is there to solve the problem. Technically rolling friction wouldn't be involved at all, correct? But then why should we know it. I can't seem to get from the 20 meters to anything else. Any suggestions?

~Phoenix

2. Sep 29, 2008

### Phoenixtears

New thoughts:

I belive that we have the initial velocities for hor. and vert.. Hori: 3 because that is the constant velocity we are given, and the vert. one is 0 because the cart is not moving vertically.

Also, can we say that we can figure out the acceleration? Because F=Ma, 7.5=380a, then a=.01974. But this can be said of neither vertical nor horizontal. So what does this acceleration represent?

I don't know where to go from here... is this information correct?

3. Oct 18, 2010

### vt.leafs.

You need to take into account the acceleration due to gravity.