Solving Spring Launcher: Find Velocity for Target Impact

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Homework Help Overview

The problem involves determining the velocity required for a spring launcher to hit a target located at a specific distance and height. The parameters provided include the height and distance of the target, as well as the angle of the launcher.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between potential energy in the spring and kinetic energy, and how to derive the necessary velocity from the given parameters. There are attempts to express the motion equations in terms of the initial velocity and to substitute variables to find a solution.

Discussion Status

The discussion has progressed with participants exploring various equations and substitutions to relate the target's height and distance to the required launch velocity. Some participants have suggested specific equations, while others are verifying the correctness of their approaches. There is no explicit consensus yet, but productive lines of reasoning are being developed.

Contextual Notes

Participants are working under the constraints of the problem's parameters, including the angle of launch and the need to backtrack from the required velocity to determine how far to pull the spring. There is an acknowledgment of potential adjustments needed in the equations based on gravitational effects.

maggiemicmuc
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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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