Solving Spring Launcher: Find Velocity for Target Impact

In summary, we need to find the velocity needed to hit the target, and then figure out how far to pull the spring.
  • #1
maggiemicmuc
5
0
[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=[tex]\sqrt{m(v^2)/k}[/tex]

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

The Attempt at a Solution



Completely lost for ideas. :confused:
 
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  • #2
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
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
Well we know: [tex]x = v_{0}\cos{\theta}t[/tex] and [tex]y = v_{0}\sin{\theta}t+\frac{1}{2}gt^2[/tex]

We can make a substitution for time and get y and x and the angle in terms of v0.
 
  • #5
Ooh okay, so the final equation would be something like:

y = vsin[tex]\theta[/tex](x/vcos[tex]\theta[/tex]) + 1/2 g (x/vcos[tex]\theta[/tex])^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
maggiemicmuc said:
Ooh okay, so the final equation would be something like:

y = vsin[tex]\theta[/tex](x/vcos[tex]\theta[/tex]) + 1/2 g (x/vcos[tex]\theta[/tex])^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...
 
  • #7
Thanks so much, that did end up working (except we had to do minus 1/2g.. instead of +1/2 g). :)
 

1. How do I determine the velocity needed for a spring launcher to hit a target?

The velocity needed for a spring launcher to hit a target can be determined by using the equation: v = √(2gh), where v is the velocity, g is the acceleration due to gravity (9.8 m/s²), and h is the height difference between the launcher and the target.

2. What are the factors that affect the velocity of a spring launcher?

The factors that affect the velocity of a spring launcher include the strength of the spring, the distance the spring is compressed, and the mass of the object being launched. Other factors such as air resistance and friction may also play a role.

3. How does the angle of the spring launcher affect the velocity?

The angle of the spring launcher can affect the velocity by changing the direction of the launch. The optimal angle for maximum velocity is typically 45 degrees, as this allows for the most efficient use of the spring's force.

4. Can I use the same formula for different spring launchers?

The same formula, v = √(2gh), can be used for different spring launchers as long as the factors that affect the velocity (such as spring strength and compression distance) remain constant. However, it is important to note that the mass of the object being launched may vary and therefore affect the velocity.

5. Are there any other methods for finding the velocity of a spring launcher?

Yes, there are other methods for finding the velocity of a spring launcher, such as using energy conservation equations or using force equations. However, the formula v = √(2gh) is the most commonly used and simplest method for calculating velocity in this scenario.

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