How Do You Derive the Range Maximizing Equation in Projectile Motion?

[Raywashere]

Homework Statement

This is for calc based physics, and the new phs prof wants us to derive this equation for a projectile motion lab.

From this range equation: R(theta) = (v_o^2cos(theta) / g) * [sin(theta) + \sqrt{sin^2(theta+ (2gh/v_o^2}]

he wants us to derive this range maximizing equation: (theta)=cot^-1\sqrt{}1 + (2gh/v_o^2)

The Attempt at a Solution

I was able to differentiate the inside of the sqrrt to 1 + 2gh/v_o^2 , because I made sin^2(theta) = to p and that just differentiated to 1, but how the heck do you get cot^-1?

also, another thing that is throwing me off is the fact that g(gravity) is a constant, so won't that just go to 0? but if that happens then the range equation can't be differentiated...right?

someone help...please.

 

[SammyS]

Homework Statement

This is for calc based physics, and the new phs prof wants us to derive this equation for a projectile motion lab.

From this range equation: R(theta) = (v_o^2cos(theta) / g) * [sin(theta) + \sqrt{sin^2(theta+ (2gh/v_o^2}]

he wants us to derive this range maximizing equation: (theta)=cot^-1\sqrt{}1 + (2gh/v_o^2)

The Attempt at a Solution

I was able to differentiate the inside of the sqrt to 1 + 2gh/v_o^2 , because I made sin^2(theta) = to p and that just differentiated to 1, but how the heck do you get cot^-1?

also, another thing that is throwing me off is the fact that g(gravity) is a constant, so won't that just go to 0? but if that happens then the range equation can't be differentiated...right?

someone help...please.

I don't mean to insult you, but your new UCLA phys. prof. probably figures that his students know how to differentiate a wide variety of functions and can handle the product rule & chain rule in particular.

BTW: if p = sin²θ , then dp/dθ ≠ 0, because p is not a constant. In fact, dp/dθ = 2(sinθ)(cosθ).

ALSO: if \theta = \cot^{-1}\sqrt{1 + (2gh/{v_0}^2)}, that means \cot(\theta) = \sqrt{1 + (2gh/{v_0}^2)}

What did you get for dR/dθ ?