# Projectile: Given V0, h, show d=(v0/g)sqrt((v0)^2-4gh)

1. Sep 12, 2008

### qamptr

1. The problem statement, all variables and given/known data
A projectile is fired from a gun (adjusted to give maximum range) with velocity $$v_{0}$$. The projectile passes through two points at a height h. The problem asks us to show that $$d=\frac{v_{0}}{g}\sqrt{v^{2}_{0}-4gh}$$
where d is the distance between the two points at height h.

2. Relevant equations
$$r=v_{0}t+\frac{1}{2}at^{2}$$
$$v=v_{0}+at$$
$$x= \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$

3. The attempt at a solution
I was able to get a quadratic function of x:
$$0=\frac{g}{v^{2}_{0}}x^{2}-x+h$$

After manipulation using the quadratic formula, all I can see is:
$$x=\frac{v^{2}_{0}}{2g}+\frac{1}{v_{0}}\sqrt{v^{2}_{0}-4gh}$$

Which just looks so close but I'm killing myself in trying to see how it is either (1) wrong or (2) able to be simplified.

EDIT: $$x=\frac{v^{2}_{0}}{2g}+\frac{1}{2gv_{0}}\sqrt{v^{2}_{0}-4gh}$$, sorry.

Help?

Last edited: Sep 12, 2008
2. Sep 12, 2008

### qamptr

Does anyone even have any suggestions? This is actually due in about an hour and a half. I'm not heartbroken or anything but I'm feeling pretty annoyed that I might not get this problem. I honestly can't see what's going wrong here. Any creative suggestions or strong nudges are totally welcome.

Thanks...

3. Sep 12, 2008

### qamptr

Alright, so, I currently have the following written on my paper:

$$x= \frac{v^{2}_{0} \pm v_{0} \sqrt{v^{2}_{0}-4gh}}{2g}$$

I can't find the correction that makes this into the formula asked for.

4. Sep 12, 2008

### qamptr

Ahhhhhhh... So, this one was staring me in the face. Done and with 20 minutes to spare.