(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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})

2. Relevant equations

x(t) = v_{0}tcosα

y(t) = v_{0}tsinα - (1/2)gt^{2}+ h

3. 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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# Homework Help: Cannon on a vertical tower

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