Range of projectile traveling through two points at same elevation

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SUMMARY

The discussion focuses on the mathematical derivation of the range of a projectile fired at an initial velocity \( V_0 \) that passes through two points at the same elevation, specifically a height \( h \) above the horizontal. The participants clarify that the relationship between height \( h \) and horizontal distance \( A \) does not equate to \( \tan(45^\circ) \). Instead, they derive \( h \) in terms of \( A \) using the equations \( A = V_0 t \cos(45^\circ) \) and \( h = V_0 t \sin(45^\circ) - \frac{1}{2}gt^2 \). The discussion emphasizes the need to express \( A \) in terms of \( h \) for maximum range calculations.

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UrbanXrisis
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"A projectile is fired with an initial velocity [tex]V_o[/tex] such that it passes through two points both a distant h above the horizontal. Show that if the gun is adjusted for a maximum range the separation of the potins is?"

my work is http://home.earthlink.net/~suburban-xrisis/projectile.pdf"

does it look correct?
 
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h/A does NOT equal tan 45. Just look at your picture to see this.
Find h in terms of A by using the eqs. A=v_0 t cos45 and
h=v_0 t sin45-(1/2)gt^2.
 
I'm not sure why they would not be equal.

Opposite/Adjacent = tan(theta)

The side opposite to the angle is h
The side Adjacent to the angle is A

so the equation should read: tan(45)=h/A

and why would I want to get h in terms of A? Dont I want A in terms of h?

I made up the variable A so I that I could get A in terms of h and subtract it from the max range.
 

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