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A projectile is fired with an initial velocity v such that it passes through two points that are both a distance h above the horizontal. Show that if the gun is adjusted for maximum range the separation of the two points is

[itex] d = v/g \sqrt{v^2 - 4gh} [/itex]

I have been struggling at this problem for much of the day. I even derived the original equations for range and max height

range, height [itex] R = v^2/g , H = v^2 / 4g [/itex]

Also we have our conservation of motion equations, and the distance function of a projectile (under the constraints of max range)

[itex] F(t) = (1/\sqrt{2} vt) \overline{i} + (1/\sqrt{2} vt - 1/2 gt^2) \overline{j} [/itex]

I attempted to use conservation of energy to find a solution to this problem, but I was having a lot of difficulty relating it to my distance function. I also tried using my distance function but I found that I could not effectively remove the time variable.

[itex] d = v/g \sqrt{v^2 - 4gh} [/itex]

## Homework Equations

I have been struggling at this problem for much of the day. I even derived the original equations for range and max height

range, height [itex] R = v^2/g , H = v^2 / 4g [/itex]

Also we have our conservation of motion equations, and the distance function of a projectile (under the constraints of max range)

[itex] F(t) = (1/\sqrt{2} vt) \overline{i} + (1/\sqrt{2} vt - 1/2 gt^2) \overline{j} [/itex]

## The Attempt at a Solution

I attempted to use conservation of energy to find a solution to this problem, but I was having a lot of difficulty relating it to my distance function. I also tried using my distance function but I found that I could not effectively remove the time variable.

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