At what throwing angle will the ball's range be maximum?

>>16715955 (OP)

>>16715956
huehuehue

>>16715955 (OP)
>baby's first newtonian physics homework

omg

>>16715959
The answer is just so obvious, you won't bother stating it. No?

>>16715955 (OP)
bout tree fiddy

Less than 45 deg if you are taller than 0.

x=cos(a)*u*t
y=-0.5*g*t^2+sin(a)*u*t+h

y=0 when flight us at end.

Solving for time

t=-sin(a)*u+(-)sqrt((sin(a)u)^2-4*-0.5gh)/(-2*0.5g)
t=(sin(a)u+(-)sqrt((sin(a)u)^2+2gh))/g
x=cos(a)u*
(sin(a)u+sqrt((sin(a)u)^2+2gh))/g

Now put d/da x= 0 to find optimum. For h=0 its 45 deg.

>>16716007
Now take drag into account to at least second-order dependence on velocity.

>>16715959
No, just discovered GORILLA.BAS

>>16715955 (OP)
at orbit angle

>>16715955 (OP)
why is the ball leaving from behind the thrower's head instead of after he's moved it forward and up to impart momentum?

>>16715964
Correct

>>16715955 (OP)
45 degrees because that's halfway between straight up and straight forward.

>>16716010
>Now take drag into account
Pride month it over, bub.

>>16715955 (OP)
A nanometer above 45 deg because there will be an infinitesimally smaller amount of air molecules and gravity if you go a bit higher.

A bigger determinate would also depend on what kind of spin you put on the ball.

It depends on a lot of things. You get more force to the ball if the angle is smaller than 45 degrees. That's how they play shot put in the olympics, they don't throw it at 45 degrees because it's harder to throw that way. Also, the fact that you're throwing in from above ground also means that the optimal angle is less than 45. The 45 degrees only applies if the ball launches from ground level.

It depends on a lot of things. You get more force to the ball if the angle is smaller than 45 degrees. That's how they play shot put in the olympics, they don't throw it at 45 degrees because it's harder to throw that way. Also, the fact that you're throwing it from a height also means that the optimal angle is less than 45. The 45 degrees only applies if the ball launches from ground level.

>>16716142
Harder for humans. What about trebuchet?

get distance in the function of angle, get angle so that the derivative of the distance is 0 I guess

>>16716007
the height doesn't matter, if all the forces involved are gravity and the throw 45deg is the answer AFAIK which is obvious to me because 45deg is the average of straight up and straight forward.

>At what throwing angle will the ball's range be maximum?
Computer engineer here. The answer is slightly over 90 degrees.

>>16716023
First of all, aiming.
Second of all, please reevaluate your understanding of the windmilling motion we call a "throw."
What determines imparted angle? Is it the angle you chose, or the release point?(If you can't answer this question, please review the concepts of "mocking" and "spoonfeeding")

>>16715955 (OP)
We don't know since the answer will depend on the consciousness of the human throwing the ball and measuring the result. This is literally the hard problem of consciousness.

>>16715955 (OP)
It depends, does the ball have an inertial dampener, a friction negater and a non-euclidian matter accelerator installed?

>>16716490
yes

>>16715955 (OP)
you have to be over 18 to post here

>>16715955 (OP)
45º

>>16716539
Wrong.

>>16715955 (OP)
stoopid you have to throw it as a straight line with infinite force.

>>16716624
That would result in it never landing, giving it a range of undefined.

>>16716629
It would eventually loop back, no? The universe is flat/shaped like a donut.

>>16716638
If it's a torus, it would eventually come back to it's start, which was the height of the hand above the ground, so it still wouldn't have hit the surface. But that's without considering how long it would take for the beam to loop around and how far the Earth will have moved since then. If it makes an infinite number of loops around the universe, perhaps eventually the Earth will at some point be in its path.

it depends

>>16716642
Infinite force means infinite speed so it will be in all locations on the path around the torus at once. According to ChatGPT, the moving Earth will bump into the ball, which is will be the same distance away from its starting point relative to the surface of the Earth as how far it started above the Earth's surface. So for infinite force, if you throw it from two meters above the surface, it will impact the surface two meters away. I have my doubts about the accuracy of ChatGPT on this one.

>>16716638
It would just continue in a straight line out the side of the torus.

Not too low below eyes

>>16716043
>>16716007
Oh look, someone beat you to it. Well, here's something that might be more your speed. What's 12^2 - 7 * 5? Pretty sure it's 685.

What is the measurement for the weight of the ball?
The weight of the ball will determine the correct throwing angle in ratio to the gravitation of the planet.
So in essence, the mass of the object thrown, in ratio, to the mass of the planet that we are throwing upon. Each will possess it's own individual gravity, and those two gravitation's interact with one another in ratio, because gravity is simply the emergence of magnetism.

>>16716624
throw at 90 degrees to exit earth

>>16717469
>See?!?!? Someone ELSE answered my homework for me!
ok timmy

>>16716010
I served in terrestrial artillery, specifically on antediluvian howitzers. we had to look up distance(angle,charge) from tables. farthest reach was still at 45 degrees for every charge, drag and velocity and Magnus effect notwithstanding (thank gawd we did not have to factor in Coriolis).