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Homework Help
Introductory Physics Homework Help
Optimizing Cannon Range on a Vertical Tower
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[QUOTE="jasonchiang97, post: 6071435, member: 545742"] [h2]Homework Statement [/h2] A cannon that is capable of firing a shell at speed v[SUB]0[/SUB] 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[SUP]2[/SUP](α) = 2(1+gh/V[SUB]0[/SUB][SUP]2[/SUP]) [h2]Homework Equations[/h2] x(t) = v[SUB]0[/SUB]tcosα y(t) = v[SUB]0[/SUB]tsinα - (1/2)gt[SUP]2[/SUP] + h [h2]The Attempt at a Solution[/h2] [FONT=PT Sans, san-serif] [SIZE=16px] First, I used x(t) = v[SUB]0[/SUB]tcosα y(t) = v[SUB]0[/SUB]tsinα - (1/2)gt[SUP]2[/SUP] + h I solved for t in y(t) and plugged it into x(t) to get x = [v[SUB]0[/SUB]cosαsinα+v[SUB]0[/SUB]cosα(v[SUB]0[/SUB][SUP]2[/SUP]sin[SUP]2[/SUP]α+2gh)[SUP]1/2[/SUP]]/g then I solved for v[SUB]0[/SUB]cosα(v[SUB]0[/SUB][SUP]2[/SUP]sin[SUP]2[/SUP]α+2gh)[SUP]1/2[/SUP] and squared both sides to get (xg - v[SUB]0[/SUB][SUP]2[/SUP]cosαsinα)/(v[SUB]0[/SUB][SUP]2[/SUP]cos[SUP]2[/SUP]α) = v[SUB]0[/SUB][SUP]2[/SUP]+2gh I expanded the left side of the equation and simplified with the right hand side to get (g[SUP]2[/SUP]/v[SUB]0[/SUB][SUP]2[/SUP])x[SUP]2[/SUP] -(2gtanα)x - 2gh = 0 Solved for x to get x = v[SUB]0[/SUB][SUP]2[/SUP]/2(sinα + √(sin[SUP]2[/SUP]α +2h/v[SUB]0[/SUB][SUP]2[/SUP])) I took the dx/dα = 0 to get the maximum angle dx/dα = 1+sinα(sin[SUP]2[/SUP]α +2h/v[SUB]0[/SUB][SUP]2[/SUP])[SUP]-1/2[/SUP] = 0 And I said that it is max when sinα = 1 so then x[SUB]max[/SUB] = v[SUB]0[/SUB][SUP]2[/SUP]/2 + v[SUB]0[/SUB][SUP]2[/SUP]/2(1+2h/v[SUB]0[/SUB][SUP]2[/SUP] I'm unsure on how to obtain cscα from this [/SIZE] [/FONT] [/QUOTE]
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Optimizing Cannon Range on a Vertical Tower
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