# Homework Help: R = Kx^2 / mg <- anybody reconise this?

1. Aug 12, 2008

### StheevilH

R = Kx^2 / mg <-- anybody reconise this?

Well..

I got this R = K x^2 / m g.. formula to explain...

where

R = range (displacement of x)
K = constant of spring (elastic)
x = extension of spring
m = mass
g = gravity

Can anybody prove this formula for me??

It says use conservation of energy and 4 formulae of projectile motion..

I'd really appreciate if anybody can prove it...

PS it is fired at angle of 45 degrees

2. Aug 12, 2008

### tiny-tim

Welcome to PF!

Hi StheevilH! Welcome to PF!
uh-uh … you have to do the work!

First write out the equation of conservation of energy, as it applies to the spring and the projectile.

Then write out as many equations for projectile motion as you can think of.

What do you get?

3. Aug 12, 2008

### StheevilH

Re: Welcome to PF!

Eh... write out as many equations for projectile motion?????

I learnt only 4 types of them though... and none of them matchs..

Before that...what is the equation of conservation of energy??

I don't get the whole thing..to begin with....so..would you be able to be more detail??

Last edited: Aug 12, 2008
4. Aug 12, 2008

### StheevilH

Re: R = Kx^2 / mg <-- anybody reconise this?

Wait.. don't worry about the above one I wrote...

I found out potential energy of the spring is 1/2 * K * x^2..

but this doesn't match with any projectile formula I got..

is x = Vx?????

or V itself? not in component form?

Gah...I'm not good at this...

5. Aug 13, 2008

### tiny-tim

Hi StheevilH!

(btw, is the spring launching a separate projectile, or is the spring itself a projectile? )

No, your PE won't fit into a projectile formula … it'll fit into a conservation of energy formula!

And conservation of energy is PE + KE = constant …

so what formula do you have for KE?

6. Aug 13, 2008

### StheevilH

Re: R = Kx^2 / mg <-- anybody reconise this?

Ah...I worked out....I think....

lol..yeah I thought PE was meant to be in projectile formula..

Eh....from what I got...is KE = PE...

1/2 * K * x^2 = 1/2 * m * v^2

Hence, v^2 = K * x^2 / m ..

Using range formula; R = Vo^2 * Sin 2 (*) / g..

R = K * x^2 / m * g....Sin 2 (45) = 1...so can be ignored...

though...

I don't get how initial velocity and the final velocity can be same?

Can you tell me how they works?

Spring itself is the projectile motion...yeah...I think...lol

Thanks if you tried!!

Oh..don't forget to reply my last question =]

7. Aug 13, 2008

### tiny-tim

Conservation of energy again … initial and final velocity are at the same height … so PE is the same … so KE must be the same!

Alternatively, just look at the curve the projectile follows … it's a parabola, and it's symmetric … so speed and angle on the way up will always be equal and opposite to speed and angle at the same height on the way down.
Sorry … what question?

8. Aug 13, 2008

### StheevilH

Re: R = Kx^2 / mg <-- anybody reconise this?

Oh..lol question...

That was pointing at the Vo being equal to V...

So... we can ignore about the directions in the calculations??

The magnitudes are same...so we can just put them as same???

and...did I actually get it that way.. @_@

9. Aug 13, 2008

### tiny-tim

No … don't forget the sin2 factor … at 45º, it just happens to be 1, but you would usually have to include it.

10. Aug 13, 2008

### StheevilH

Re: R = Kx^2 / mg <-- anybody reconise this?

Roger, copy that

(Y)