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

  • Thread starter qamptr
  • Start date
  • #1
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Homework Statement


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

Homework Equations


[tex]r=v_{0}t+\frac{1}{2}at^{2}[/tex]
[tex]v=v_{0}+at[/tex]
[tex]x= \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex]


The Attempt at a Solution


I was able to get a quadratic function of x:
[tex]0=\frac{g}{v^{2}_{0}}x^{2}-x+h[/tex]

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

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: [tex]x=\frac{v^{2}_{0}}{2g}+\frac{1}{2gv_{0}}\sqrt{v^{2}_{0}-4gh}[/tex], sorry.

Help?
 
Last edited:

Answers and Replies

  • #2
10
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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
10
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Alright, so, I currently have the following written on my paper:

[tex]x= \frac{v^{2}_{0} \pm v_{0} \sqrt{v^{2}_{0}-4gh}}{2g} [/tex]

I can't find the correction that makes this into the formula asked for.
 
  • #4
10
0
Ahhhhhhh... So, this one was staring me in the face. Done and with 20 minutes to spare.
 

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