It’s still a good exercise in problem solving I’d say (although I do question why they’d make this a homework assignment).
They didn’t
I say it isn’t at all, I get learning shit like riemann sums and that getting thrown out the window later because you built upon that as a foundation while something like this method I’ve never heard and many like me haven’t clearly is nigh useless since an easy to learn method for the actual way to do it already exists.
American education spending too much time on useless shit instead of relevant studies
No need to be rude, that’s a fairly reasonable thing to say if you haven’t learned trig yet. It is a very loaded subject matter so from the outside it is a lot to take in.
That’s what I’m saying bro they teach 1st AND 2nd graders “sight words” and so basically THEY CAN’T READ until 4th grade like wth are u doing just teach them to read
i understand that you’re new here but that guy posted the most diabolically bad fourth rate dogshit suggestion of 2023 and became the #1 most wanted on these streets 
bec everyone thought its for pvp
I don’t really care I just want to help him learn some cool math.
Btw @opticalcord Z is equal to 28.8 for the bunny slipper question.
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Just an aside but I love riemann sums to death, very useful for integrating over discrete values (such as deltatime in program development).
1 2 3 5 8 13 21 34
what comes after 34
55 
…35
35 comes after 34
isnt trig 10th grade though
mb gang i forgot how to count
9 + 10
I am charging a poison tooth dagger throw attack. Assume the projectile travels at 150 units per second. My position is at (0, 100). My opponent is falling and they start at (180, 160). Don’t give a shit about the Z axis for now. Assume gravity acceleration is 28 units per second. No terminal velocity.
What angle should I aim the poison tooth dagger so that it hits my opponent (or is it impossible)?
this is a physics question
answer is charging dagger throw and aiming at where the opponent lands (0, 160)
no need for mathing when you can just get your money up (hbe installed 
)
How fast is the opponent falling initially?
28 per second (just started)
Lets first derive the formulas for the acceleration, velocity, and positions of both the player and the opponent.
For the Y direction:
The equation for acceleration is given as A(x) = 28.
The equation for velocity is the derivative of acceleration, and will thus be V(x) = v_i + ax.
The equation for position is the derivative of velocity, and will thus be P(x) = p_i + v_i * x + (ax^2)/2.
For the X direction:
Velocity remains constant, so the velocity equation is V(x) = v_j.
Position is the derivative, so the position equation is P(x) = p_j + v_j * x.
Note:
These equations work for both the dagger and the opponent, however the variables will be separate between the two. To avoid confusion, I will put the initial of the object being used in the base of the object.
(ie. the initial velocity of the dagger would become v_di, rather than the general form v_i.)
Finding the variables
We know that the initial position for the dagger is 100 in the Y direction, so p_di = 100.
Getting the velocity is a bit trickier, since the magnitude of both v_di and v_dj must equal 150. We will have to use some trig to find it out.
v_di and v_dj are both found by using trig functions on the magnitude of 150. Here’s a little image to help explain why:
Sine governs the Y direction, so it would make sense that v_di = 150*sine(θ).
Cosine governs the X direction, so it would make sense that v_dj = 150*cosine(θ).
For the opponent, their initial positions are 180 on the X direction (p_oj = 180) and 160 on the Y direction (p_oi = 160).
Their Y velocity is also given as 28 u/s, so v_oi = 28.
Solving
Now we need to solve the system of equations to find θ.
The first equation for the Y direction would be the position of the dagger equaling the position of the opponent. That can be found like so:
p_di + v_di * x + (ax^2)/2 = p_oi + v_oi * x + (ax^2)/2
plugging in our variables we get the equation:
100 + 150 * sine(θ) * x + 14 * x^2 = 160 + 28 * x + 14 * x^2
We can eliminate the 14 * x^2 on both sides to get:
100 + 150 * sine(θ) * x = 160 + 28 * x
Moving 130 over to the left will give us this:
30 + 150 * sine(θ) * x = 30 + 28 * x
Note that in order for this equation to be true, 150 * sine(θ) MUST be equal to 28. We can solve for it like so:
150 * sine(θ) = 28
Move 150 over to the right hand side.
sine(θ) = 28/150
use arcsine to isolate the θ
θ = arcsine(28/150)
This gives us an angle θ of 0.188-ish radians, or around 10.76 degrees.
Lets check if this is true by looking at the X direction.
Setting up the equation and we get:
p_dj + 150 *cosine(θ) * x = p_oj + v_oj * x
Plugging in variables gives us:
150 *cosine(θ) * x = 180
Lets divide both sides by 150:
cosine(θ) * x = 18/15
Inconclusive, any non-zero, non-negative angle would be able to satisfy that equation.
Conclusion
idk