Height of a projectile at two points?

[Vitani11]

Homework Statement

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

d = (v/g)(√v²-4gh)

Homework Equations

y = -gt²/2 + vsinθt
x = vcosθt

for max range R = v²/g
θ = π/4

The Attempt at a Solution

substituting theta into the two equations gives me
h = -gt²/2 + vt/√2

d = (v/√2)t

I replaced y with h to find the two times at which the ball is at h (both going up and down), then I used those times I found in the d equation to find the distance at both points at both times, and then took the difference between "d1" and "d2" where d2 is the point after max height and d1 is the point before. Is this not the right method? I get that t = h and t = stuff from the h equation... and t = h is obviously not true. Is that the issue?

 

[haruspex]

Is this not the right method?

Looks right.

I get that t = h

Then you are making an algebraic error. You can use dimensional analysis to see where it goes wrong. If you cannot spot it, please post all your working.

 

[Vitani11]

h = -gt²/2 + vt/√2 to begin. I pull a t out and it becomes h = t(-gt/2+v/√2) so h = t and -gt/2+v/√2 = h

:/

 

[haruspex]

h = t(-gt/2+v/√2) so h = t

By what logic?

 

[Cutter Ketch]

h = -gt²/2 + vt/√2 to begin. I pull a t out and it becomes h = t(-gt/2+v/√2) so h = t and -gt/2+v/√2 = h

:/

That looks suspiciously like what you would do if you were trying to find the zeros of an equation. i.e. t(-gt/2+v/√2) = 0 so t = 0 or
(-gt/2+v/√2) =0. That isn't at all correct for h.

 

[Vitani11]

Finished. Thank you.