Suppose you throw a ball off the edge of a cliff of height h...

Isn't this only for flat ground? If you're elevated there's going to be some extra distance. The answer is definitely below 45°. It depends on the height difference.

Yes it is. But if it achieves max distance on level ground, wouldn't it keep achieving this superiority over the other angles if there were a greater fall since horizontal velocity remains constant?

Also, a greater angle affords more time since sin(theta) would be higher = more time spent vertically as well.

I just wish I could prove it.

Try this simulator: walter-fendt.de/html5/phen/projectile_en.htm

Oh, this simulator actually shows that 45 degrees isn't the optimal angle when there's an initial height greater than zero...

Turns out you're right, and the people saying it's 45 degrees are wrong. Using the simulator from given an initial height of 100m, mass 1kg, and gravitational acceleration of 9.81 the optimal angle is given as 17-18 degrees as predicted in my original answer from

plugged 45 degrees as the angle; turns out a height of 0 is needed for 45 degrees to be the optimal angle for max range.

Imagine these arcs continuing if you moved the ground down 20 feet. Which one would move the farthest horizontally?

I'd imagine the 30 degree one, since it has a greater x component of velocity.

Depends on whether the 30 degree ball could catch up to the 45 degree one given the time.

>walter-fendt.de/html5/phen/projectile_en.htm
i fucking hate when "educational" "simulators" put arbitrary limits on value range

I think the greater the height the lower the angle. If the initial height is really high, close to infinity, then the ball traveling close to 0 degrees in a straight line horizontally would give it the max distance.

You're asking the wrong question if you want to find out how to throw the ball so that it travels the it's maximum horizontal distance. You should be asking how high up you need to go and how fast you need to throw the ball so that it enters a stable orbit and thus travels an infinite horizontal distance.

Infinity is the maximum.

Pi/4 radians

It depends on initial height. If you use g = 9.81 m/s^2 and [math]V_0[/math] = 10 m/s you can see some horizontal distance vs launch angle (in rad) plotted here, with 3 sample initial heights. Each maximum occurs at a different angle. If you're curious, the equation for horizontal distance ends up being this (with x and h in meters and [math]V_0[/math] in m/s): [eqn]x(\theta,h,V_0,g)=V_0cos(\theta) (\frac{V_0 sin(\theta)}{g} +
\sqrt{\frac{2gh + {V_0}^{2} sin(\theta)^2}{g^{2}}})[/eqn]

Neat. What did you use to make the graph?

Forgot to add that the optimal angle is [math]\theta = {cos^{-1}}(\frac{\sqrt{2gh+{V_0}^2}}{\sqrt{2gh+2{V_0}^2}})[/math]

Mathematica

Could you graph something that would have that optimal angle as well as the angle found in to see if they return the same results?

They're equal

Ah, good. What program did you use? Is that matlab?

Mathematica. It's handy but it's pretty pricey.

Gonna have to see about getting it when i have $$$.

I'm pretty sure it's 45 degrees

Wtf I was taught wrong?

Niggers wtf, the 45 degree shit only works when you are on a flat ground. You all failed basic mechanics.