Derive optimal anlge for maximum range (Projectile Motion)

[MohammadG]

Homework Statement

Hey,

For an assignment I need to derive the optimal angle for maximum range (derive Equation 2 below). I know how to derive equation 1 but I need to derive the second equation so I can substitute h = 0, and h = 1, into it, to show the optimal angle for maximum range for bi-level and uni-level.

I have a book that explains it but I don't understand it at all. Any help is really appreciated.

Thank you!

Homework Equations

Equation 1:
Rmax = u^2 / g √(1 + 2gh/u^2)

Equation 2:
tan θ = 1 / √ (1 + 2gh / u^2)

u = initial velocity
g = grav. accel.
h = height

The Attempt at a Solution

One method I have tried is finding the derivative of Rmax, setting it to zero and solving for theta (to find the minimum) which does give me 45 degrees but I cannot apply this to bi-level projection.

 

[Screwdriver]

The way you've written the fractions is somewhat ambiguous. It's hard to say whether you meant that $R_{\text{max}}=\frac{u^2}{g}\frac{1}{\sqrt{1+\frac{2gh}{u^2}}}$ or $R_{\text{max}}=\frac{u^2}{g}\sqrt{1+\frac{2gh}{u^2}}$ (especially since the term in the radical is dimensionless), but I've realized that it's the second one - the moral of the story is to use parentheses .

In this case, one method is to use the standard $x(t)$ and $y(t)$ to find $y(x)$ and sub in $y(R)=0$ to get an equation that defines $R$ implicitly as a function of $\theta$. Knowing this, you can find $\tan(\theta_{\text{max}})$ as a function of $R_{\text{max}}$, which you said you already found.