Optimal angle for a projectile from height h

[RMalayappan]

I'm following Professor Ramamurti Shankar's video lecture series for Fundamentals of Physics I on Open Yale and I'm hung up on the last question on the second problem set, of which there is no mention on the solution set.
http://oyc.yale.edu/physics/phys-200/lecture-4
The question asks, "Show that if a projectile is shot from a height h with speed v₀ the maximum range obtains for for launch angle [tex]\theta = ArcTan(\frac{v_0}{\sqrt(v_0^2+2gh)})[/tex]

I tried this by starting with y=h+v₀sinθt-1/2gt² and x=v₀cosθt. I solved for t in the y equation using the quadratic formula and got the final time t_f = (v₀sinθ+√(2gh+(v₀sinθ)²))/g and I substituted it into the x equation and tried differentiating with respect to θ and I got some monstrosity that looks like it won't yield the answer and looks like too much work anyways for the problem. The problem is that everywhere I find this √(2gh+(v₀sinθ)²) term, which looks like it should be replaced with a √(2gh+v₀²) term to get the intended result. Is there an easier approach that I can use to obtain the answer I'm looking for?

 

[Dr. Courtney]

Try and simplify xf after you plug in tf by using trig identities. Reducing the amount of times theta occurs in xf will make differentiating much simplier.

 

[RMalayappan]

I have, but I still keep having to do a product rule with a square root term that ends up becoming messy.

 

[haruspex]

An easier start would be to use the x direction equation to express t in terms of the angle then substitute for t in the y equation.
To assist any further, we need to see your working.

 

[RMalayappan]

I don't think I can substitute t into the y equation since I'm trying to maximize x and I need to solve for t when y=0.
Here is my work so far:
[itex]y(t)= h+v_0sin(\theta)t -\frac{1}{2}gt^2=0[/itex]

[itex]t= \frac{-v_0sin(\theta)\pm \sqrt{v_0^2sin^2(\theta) +2gh}}{-g}[/itex] and disregarding the negative solution:

[itex]t_f= \frac{v_0sin(\theta) +\sqrt{v_0^2sin^2(\theta) +2gh}}{g}[/itex]

Substituting into the x equation [itex]x=v_0cos(\theta)t[/itex]:

[itex]x_f(\theta)= v_0cos(\theta) \frac{v_0sin(\theta) +\sqrt{v_0^2sin^2(\theta) +2gh}}{g}[/itex]
Differentiating [itex]x_f[/itex] with respect to [itex]\theta[/itex] yields a complicated expression that looks like hell to solve. After simplification, it becomes:

[itex]\frac{dx_f}{d\theta} =0= \frac{v_0^2}{g} (cos(2\theta) +\frac{sin(2\theta)cos(\theta)}{\sqrt{sin^2(\theta) +\frac{2gh}{v_0^2}}} -sin(\theta) \sqrt{sin^2(\theta)+ \frac{2gh}{v_0^2}})[/itex]

and there I decided to look for other options. I'm going to try to use Lagrange multipliers to solve for the maximum of [itex]x(\theta, t)[/itex] along the constraint [itex]y(\theta, t) = 0[/itex] using [itex]\nabla x = \lambda \nabla y[/itex]

 

[haruspex]

I don't think I can substitute t into the y equation since I'm trying to maximize x and I need to solve for t when y=0.

Why do you need to solve for t? Substitute for t as I suggested, set y=0 and differentiate wrt theta to find max x.

 

[RMalayappan]

The issue I found with that is that h drops out of the equation once y=0 with x substituted is differentiated implicitly with respect to theta. I managed to solve the equation with Lagrange multipliers with the following:

[itex]\nabla x = \lambda \nabla y[/itex] so [itex]\frac{\partial x}{\partial \theta} = \lambda \frac{\partial y}{\partial \theta}[/itex] and [itex]\frac{\partial x}{\partial t} = \lambda \frac{\partial y}{\partial t}[/itex]
From [itex]\frac{\partial x}{\partial \theta} = \lambda \frac{\partial y}{\partial \theta}[/itex], [itex]-v_0sin(\theta)t = \lambda v_0cos(\theta)t[/itex] so [itex] \lambda = -tan(\theta)[/itex]
From [itex]\frac{\partial x}{\partial t} = \lambda \frac{\partial y}{\partial t}[/itex], [itex]v_0cos(\theta) = \lambda (v_0sin(\theta)-gt)[/itex] and, by substituting [itex] \lambda [/itex] in, [itex]v_0cos(\theta) = -tan(\theta)(v_0sin(\theta)-gt)[/itex]
After some algebra: [itex]v_0(cos^2(\theta)+sin^2(\theta)) = gtsin(\theta)[/itex] so [itex]t = \frac{v_0}{gsin(\theta)}[/itex]
Substituting into y=0, simplifying and multiplying through by 2g yields [itex]0= 2gh+2v_0^2-v_0^2csc^2(\theta)[/itex]
Using the Pythagorean identity and rearranging, [itex]v_0^2cot^2(\theta)=2gh+v_0^2[/itex] so [itex]cot^2(\theta)= \frac{2gh+v_0^2}{v_0^2}[/itex]
Here the desired result is obvious: [itex]\theta = arctan(\frac{v_0}{\sqrt{2gh+v_0^2}})[/itex]

 

[haruspex]

The issue I found with that is that h drops out of the equation once y=0 with x substituted is differentiated implicitly with respect to theta.

That is true, but it's ok. You still have the undifferentiated form of the equation with h in it. Both equations are true. Treat them as a pair of simultaneous equations and solve. It is no more difficult than your method above.