(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

[FONT=PT Sans, san-serif]

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

[/FONT]

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Cannon on a vertical tower

Have something to add?

**Physics Forums | Science Articles, Homework Help, Discussion**