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## Homework Statement

A cannon that is capable of firing a shell at speed v

_{0}is mounted on a vertical tower of height h that overlooks a level plain below.

Show that the elevation angle α at which the cannon must be set to achieve maximum range is given by the expression

csc

^{2}(α) = 2(1+gh/V

_{0}

^{2})

## Homework Equations

x(t) = v

_{0}tcosα

y(t) = v

_{0}tsinα - (1/2)gt

^{2}+ h

## The Attempt at a Solution

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First, I used

x(t) = v

_{0}tcosα

y(t) = v

_{0}tsinα - (1/2)gt

^{2}+ h

I solved for t in y(t) and plugged it into x(t) to get

x = [v

_{0}cosαsinα+v

_{0}cosα(v

_{0}

^{2}sin

^{2}α+2gh)

^{1/2}]/g

then I solved for v

_{0}cosα(v

_{0}

^{2}sin

^{2}α+2gh)

^{1/2}

and squared both sides to get

(xg - v

_{0}

^{2}cosαsinα)/(v

_{0}

^{2}cos

^{2}α) = v

_{0}

^{2}+2gh

I expanded the left side of the equation and simplified with the right hand side to get

(g

^{2}/v

_{0}

^{2})x

^{2}-(2gtanα)x - 2gh = 0

Solved for x to get

x = v

_{0}

^{2}/2(sinα + √(sin

^{2}α +2h/v

_{0}

^{2}))

I took the dx/dα = 0 to get the maximum angle

dx/dα = 1+sinα(sin

^{2}α +2h/v

_{0}

^{2})

^{-1/2}= 0

And I said that it is max when sinα = 1 so then

x

_{max}= v

_{0}

^{2}/2 + v

_{0}

^{2}/2(1+2h/v

_{0}

^{2}

I'm unsure on how to obtain cscα from this

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