Predicting X ( point of impact )

[King_Cool]

Hi
I Have a moving Ball Object.
This is known about the Ball:
- Its starting coordinates ( X and Y )
- Its Angle ( 360 )
- Its speed ( pixels moved in direction of Angle per Frame )

Eveything above Y possition 380 in the Frame is ground level.

Given this data, how would i be able to make up a formula to predict at what X coordinate along the ground level the Ball will be impacting?
If i knew where to start with this i woulden't ask :P

[Konidias]

I'm pretty terrible at math so I can't really come up with the formula for that on the spot...

My trick would be to create an invisible ball object identical to the visible one, and then have it hit the target quickly to determine where the visible ball object would hit.

[danjo]

its just a trig question;
heres a link to demonstrate how to find the answer you are looking for: http://www.mathwarehouse.com/trigono...-practice3.php

in your case, you just need to know how far up from the ground the ball is (ground Y - ball Y)
and the angle it travels.
from those 2 pieces of info, you can determine how far it will travel (x) before reaching the ground. add or subtract that amount from the balls original X will give you the X point on the ground.

[MuddyMole]

In other words...
X - ((380 - Y) * (cos(angle) / sin(angle)))

[King_Cool]

OK so..

Ball Xpos - Ball Y distace from Ground * (cos(angle) / sin(angle)

...
(cos(angle) / sin(angle)???

Why do i do this?

[MuddyMole]

To get the gradient of the straight line along which the ball is moving. Obviously the ball's speed is irrelevant.

For every 1 pixel the ball moves down, it moves "cos(angle) / (-1 * sin(angle))" pixels to the right (the -1 part is because of MMF2's inverted y-axis). You know how many pixels down the object needs to move (obviously it's 380 - Y), so you can just multiply the two together to find out how far to the right the ball needs to move. And once you know how far right it's going to move from its starting point, you simply need to add the X position where it started.

If you subtract from X instead of adding to it, you don't need the "-1 * " part - it just makes the formula a bit neater.

[King_Cool]

Awesome it works!
Thank you so much MuddyMole!