1st year Physics Question - Difficult 2D-motion

Hello. I am a 1st year Engineering Physics student and I am having a bit of a problem with one of the questions. I was wondering if anyone here could possibly help me, for instance show me what to do to be able to solve it. Here is the question:

"A cannon can project a cannonball from its barrel with a certain muzzle speed v[SUB]0[/SUB]. Ignoring the effects of air resistance, what formula expresses the distance the cannonball travels before it reaches the ground as a function of the angle 'theta' that the barrel makes with the ground? Unlike the projectile treated in Section 3-4, the cannonball is fired from the edge of a cliff of height h[SUB]0[/SUB] above the level plain at which it is aimed. Show that the angle 'theta' that gives the largest horizontal range is given by sin[SUP]2[/SUP]'theta' = v[SUB]0[/SUB][SUP]2[/SUP]/2(v[SUB]o[/SUB][SUP]2[/SUP] + 2gh[SUB]0[/SUB]). [Hint: In calculus, you learn that a function of a variable such as 'theta' has a maximum (or a minimum) at an angle 'theta[SUB]0[/SUB]', the angle for which the derivative of the function with respect to 'theta' is zero."

I finished the first part of the question, but am having a huge difficulty trying to get the second part. What I tried to do was split the function into components of y and x, where both are related to theta and time. I then had the y component = 0 to use the quadratic formula to find t, and then plugged t into x to relate x with respect to the angle. I then took the derivative (huge, long mother-f) and attempted to set it to 0 and simplify. But, I mess up. Any help from the smart people of Southperry is appreciated!

The way I would do it is finding the formula for distance in x (Part 1 of your problem which you said you already got), derive it with respect to Theta, set to zero and see if the resultant equation equals the equation given to you.
But yeah, it's a lot of crappy algebra and calculus.

For the second part...The theta that gives the largest angle is always 45 degrees, remember that to heart. The ones that fail are 90 and 0 degrees.

That formula is gay to be honest, use the formula

R = (Vo^2)(sin2Theta)/g

That is, the Range equals the (initial velocity squared) times (the sin of double the angle) all divided by g. The perfect angle will be 45 degrees because double that = sin 90 and the sin of 90 = 1, anything besides 45 degrees will yield a fraction which will make the range shorter.

I wish I had more time to draw you a diagram but I have to go to school.

Unfortunately, that wouldn't be correct for the second part. It is only 45 degrees if it is on a flat surface, say if h is 0. If you calculate it with an initial height of 1, for instance, and calculate the distance when you use 44, 45, and 46 degrees, you would find you get a furthur distance with 44.

Oh I see. I get it... I took the derivative of sin^2 theta with respect to theta and set it equal to zero, and theta did turn out to be zero hmmm...This question is tricky... Still thinking lol.

first thing you need to ask is "how long is it in the air?", you've got that down.

h_0 + v_0 sin(theta) t - 1/2 g t^2 = 0

the 3 terms, in order: initial height, velocity in y direction * t, acceleration * t^2
solve for t. once you have t, figure out how far it goes.

v_0 cos(theta) t = R

simply velocity in x direction * air time

sub in your non-physically-bad solution for t from the y equation to the x equation. now you have R(theta), starting from any height. take the derivative w.r.t. theta and set dR/dtheta = 0. solve for theta. or rather, all of the theta's should be inside sin's and cos's, so massage those to get your sin^2(theta).

that's how i would do this problem. it sounds like you were on this track though. so check your derivative and t solution?

Darn. I checked both my t value and my derivative with wolfram alpha, and they are correct. It appears as if it is some sort of simplification error on my part. Tedious. I'll continue working on it, and let you guys know how it goes.

Figured it out. Apparently the textbook was wrong, and the answer was supposed to be sin^2(x)=(v^2)/2(v^2+gh)

Off by a darn 2. Thanks for the help, everyone!