Solving Spring Launcher: Find Velocity for Target Impact

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To determine the velocity needed for a spring launcher to hit a target at a specific distance and height, the key equations involve the potential energy of the spring and projectile motion. The potential energy of the spring converts to kinetic energy, allowing for the calculation of velocity based on the spring's compression. The trajectory equations for horizontal and vertical motion can be combined to express the height and distance in terms of the launch velocity and angle. After substituting the time variable, the final equation can be rearranged to solve for the required velocity. Adjustments to the gravitational term in the equation are necessary for accurate results.
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[SOLVED] Spring launcher

Homework Statement



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.

Homework Equations



F=kx

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

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

The Attempt at a Solution



Completely lost for ideas. :confused:
 
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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
 
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.
 
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.
 
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?
 
maggiemicmuc said:
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?

Yeah, that's pretty much what I was thinking...
 
Thanks so much, that did end up working (except we had to do minus 1/2g.. instead of +1/2 g). :)
 
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