Optimal launch angle on an uneven ground

[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 ?

 

[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.