# Optimal launch angle on an uneven ground

1. Sep 5, 2014

### throneoo

Hi guys,

I was working on a formula for the optimal angle for simple point-like projectiles free-falling under a uniform g-field but I got stuck.

after a simple derivation , I obtained the following equation:

R=(vcos(@)/g)(vsin(@)+((vsin(@))^2+2gh)^0.5)

where R , @ , h are respectively the range as it reaches the ground , launch angle and initial height

I then took a derivative w.r.t. @ and set dR/d@ =0 , and obtained the following equation:

vsin^2(@)+ksin(@)=vcos^2(@)+(v^2sin(@)cos^2(@))/k

where k=((vsin(@))^2+2gh)^0.5

I'm quite certain that I got everything right as when h=0 , the equation reduces to tan(@)^2=1 and immediately yields @=pi/4

but with a non-zero 'h' I fail to see how I can solve for @ in terms of the parameters via normal algebraic methods..

How else can I work on it ?

2. Sep 6, 2014

### voko

I tried that a bit differently, and obtained the following: $$R \propto \alpha \left(\beta + \sqrt {\beta^2 + K} \right),$$ where $$K = {2gh \over v^2}$$ and $\alpha, \beta$ are the cosine and sine of the launch angle, because of which we have $$\alpha^2 + \beta^2 = 1.$$ Using Lagrange's multiplier method, let $$F(\alpha, \beta, \lambda) = \alpha \left(\beta + \sqrt {\beta^2 + K} \right) + \lambda \left(\alpha^2 + \beta^2 - 1\right) .$$ Then $${\partial F \over \partial \alpha} = \left(\beta + \sqrt {\beta^2 + K} \right) + 2 \lambda \alpha = 0,$$ $${\partial F \over \partial \beta} = \alpha \left(1 + {\beta \over \sqrt {\beta^2 + K} } \right) + 2 \lambda \beta = 0.$$ The latter two equations yield $$2 \lambda \alpha^2 = 2 \lambda \beta \sqrt {\beta^2 + K},$$ which, combined with $\alpha^2 + \beta^2 = 1$, results in $$1 - \beta^2 = \beta \sqrt {\beta^2 + K}.$$ Subsequent steps should be trivial.