Spring gun problem can't find start

[JWDavid]

Homework Statement

spring-loaded gun on tabletop. Target box 2.20m horizontally from edge of table on floor.
Shot 1 - spring compressed 1.10 cm - falls 27 cm short.
How far should the spring be compressed to hit the target with shot 2?

Homework Equations

GPE1 + SPE1 + KE1 = GPE2 + SPE2 + KE2<br>
0 + k(x²)/2 + 0 = mgh + 0 + 0

The Attempt at a Solution

I've been beating myself up trying to solve this one and the next one I'll post and I don't know where to start. <p>
I tried:<br>
to use a range solution (V²Sin(2Theta)/g) and cut it in half to get the initial velocity so I could attempt to figure out KE (mV^{2/2 - but don't have an angle<p>
to figure out time of flight - but don't have the height of the table<p>
I don't have a spring constant for it. And I just don't think I should be using proportions to determine the answer, not this late in a physics (w/calculus) class.<p>
I don't know where to start, and btw it didn't give me mass either.<P?
Thanks in advance
o}

 

[nasu]

See "Spring and projectile problem" on Nov 30. Same problem, already discussed on the forum.

 

[LowlyPion]

Homework Statement

spring-loaded gun on tabletop. Target box 2.20m horizontally from edge of table on floor.
Shot 1 - spring compressed 1.10 cm - falls 27 cm short.
How far should the spring be compressed to hit the target with shot 2?

Homework Equations

GPE1 + SPE1 + KE1 = GPE2 + SPE2 + KE2<br>
0 + k(x²)/2 + 0 = mgh + 0 + 0

The Attempt at a Solution

I've been beating myself up trying to solve this one and the next one I'll post and I don't know where to start. <p>
I tried:<br>
to use a range solution (V²Sin(2Theta)/g) and cut it in half to get the initial velocity so I could attempt to figure out KE (mV^{2/2 - but don't have an angle<p>
to figure out time of flight - but don't have the height of the table<p>
I don't have a spring constant for it. And I just don't think I should be using proportions to determine the answer, not this late in a physics (w/calculus) class.<p>
I don't know where to start, and btw it didn't give me mass either.<P?
Thanks in advance
o}

<sup>Whatever the height of the table you know that the horizontal velocity times the time to fall is short i.e. too slow. So to get to the proper distance the horizontal speed needs to be increased to 220/193 of whatever speed it comes off the top of the table.

Working backward, the v will come from the kinetic energy that came from the potential energy of the spring in the gun. Since the mass and the spring constant are invariant that means that the x is proportional to the v.

So ... that means that if the x is 220/193 of the 1.1 cm you might have a better chance?</sup>

 

[JWDavid]

Thank you both, I understand LowlyPion's answer better, but I think I learned more from figuring out how to get where nasu's answer pointed. Thanks again.

 

[LowlyPion]

Thank you both, I understand LowlyPion's answer better, but I think I learned more from figuring out how to get where nasu's answer pointed. Thanks again.

I just read the other thread. It can be found here. (Not Nov 30, Oct 30)
https://www.physicsforums.com/showthread.php?t=268009

Interesting ... but a long way around the barn I think.