# Spring launcher

1. Dec 16, 2007

### maggiemicmuc

[SOLVED] Spring launcher

1. The problem statement, all variables and given/known data

A spring is launched to hit a target. The target is at a certain distance and height. How do you find the velocity needed to hit the target? The height and distance of the target, and the angle of the launcher are given.

2. Relevant equations

F=kx

x=$$\sqrt{m(v^2)/k}$$

v=$$\sqrt{gx\Delta/(sin2\theta)}$$

3. The attempt at a solution

Completely lost for ideas.

2. Dec 16, 2007

### Feldoh

A spring has potential energy equal to .5kx^2, when the spring is relaxed the energy changes to kinetic energy .5mv^2, so you can figure out the velocity

3. Dec 16, 2007

### maggiemicmuc

Thanks! But we've figured that out; we know how to figure out the velocity when we know how far back we pull the spring (x), but what we don't know is how to find which velocity we'd need to get the spring to hit the target. Once we get that velocity, we'd work back and figure out how far back to pull the spring.

So we have the height of the target, the distance it is from the launcher, and the angle the launcher is set to. Then we need to know which velocity it would take to get the spring to hit the target, and then figure out how far to pull the spring. We just don't know how to figure out that velocity.

4. Dec 16, 2007

### Feldoh

Well we know: $$x = v_{0}\cos{\theta}t$$ and $$y = v_{0}\sin{\theta}t+\frac{1}{2}gt^2$$

We can make a substitution for time and get y and x and the angle in terms of v0.

5. Dec 16, 2007

### maggiemicmuc

Ooh okay, so the final equation would be something like:

y = vsin$$\theta$$(x/vcos$$\theta$$) + 1/2 g (x/vcos$$\theta$$)^2

Is that right? And then you re-arange to find v, and sub in y and x for the height and distance of the target?

6. Dec 16, 2007

### Feldoh

Yeah, that's pretty much what I was thinking...

7. Dec 17, 2007

### maggiemicmuc

Thanks so much, that did end up working (except we had to do minus 1/2g.. instead of +1/2 g). :)