math problem

i have a ball with init speed V0 shooting at angle θ. So it's components in x and y axis will be:

V0y = V0*Sinθ
V0x = V0*Cosθ

And the formula for distance traveled in x axis is:

Sx = V0x*t
therefore
t = Sx/V0x

And formula for distance traveled in y axis:

Sy = V0y*t + 0.5*a*t^2
Replacing t will give us:
Sy = V0y*(Sx/V0x) + 0.5*a*(Sx/V0x)^2
This is equal to:
Sy = V0*Sinθ*(Sx/V0*Cosθ) + 0.5*a*(Sx/V0*Cosθ)^2

Here comes the problem:
I know the value for these variables:
Sy, Sx, V0 and a
How do i find initial shooting angle θ?
Remember Sy isnt 0 it can be any real value.

I know this will have a max range, but i need the formula to find this max range. example if Sy=0 i will get at the end:

Sx = -(v^2*sin2θ)/a
Here i know the max is when sin2θ = 1
But when Sy is not 0, i cant get this, cus then Cosθ and Sinθ exists at the same time... my calculus skill is not good enough to unite them to only sin or only cos

If movement in Y axis is defined by the formula: y(x) = Ax^2 + Bx +C

then θ = Atan(B)

can u send me the website or somewhere to prove this?

Wiki is your friend ^^
or if you distrust wikipedia try this: nice alternative

amazing :D

42

tdr cant expect u being here. sorry i havent posted many paintings of mine XD

Well, there's not website, in fact it was the result taht I got when I did my calculations to develop projectile movement.

Assuming you know calculus and derivatives, you must know that dy/dx represent how the Y distance varies with X distance. Or in this case, for your convenience, the function that represents the slope of a line tangent to any point of the original function.

So if your function is in the way y(x) = Ax^2 + Bx + C, then its derivative will have this form: dy/dx(x) = 2Ax + B

for x = 0, dy/dx(0) = 2A*0 + B = B

and because dy/dx is the slope of the line tangent, then the angle is obtained by the Atan... so... teta = Atan(B)

I hope this explanation were enough for you :)

Whoa, that's kewl moyack. I've been casually learning differential calculus, and never really thought about using it that way! Thanks!

I get u moyack, but this wasnt wat i wanted to find out... because here you just figure out the angle for the parabol. But problem is that ur formula only finds angle dependent on the x distance, not taken into account of y distance...
But ur equation is a good start for finding the futher equations for when this y(x) cross a point (x,y), and when we get total equation, we can find this correct angle

and from wiki link i found this: Such a nice tool

Moyack's observation is general for any parabola. The slope at which you want to shoot at is given by "B". It looks like you solved for A,B,C given a particular set of boundary conditions (trying to shoot at a point (x,y) relative to (0,0)) and then plugged it into his formula. So that we can reference this thread and that image in particular in the future please clarify if this is incorrect.

is "you" me or moyack?
Anyway thanks for the replies all :)

MaD[Lion]

i didnt plug into his formula i think, i just took tat formula from wikipedia :P