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    <p>I'm assuming that you want the ball to hit that specific point (200,200) at the apex of its path. Well, my physics is a bit rusty, but this is what I've thrown together:</p> <p><code>v_y = square_root(2*g*y)</code>,</p> <p>where g is a positive number reflecting the acceleration due to gravity, and y being how high you want to go (200 in this case).</p> <p><code>v_x = (x*g) / v_y</code>,</p> <p>where x is how far in the x direction you want to go (200 in this case), g is as before, and Vy is the answer we got in the previous equation.</p> <p>These equations remove the need for an angle. However, if you'd rather have the velocity + angle, that's simple:</p> <p><code>v0 = square_root(v_x^2 + v_y^2)</code></p> <p>and</p> <p><code>angle = arctan(v_y / v_x)</code>.</p> <p>Here is the derivation, if you're interested:</p> <p><code>(1/2)at^2 + v_yt + 0 = y</code></p> <p><code>(1/2)at^2 + v_yt - y = 0</code></p> <p>by quadratic formula,</p> <p><code>t = (-v_y +/- square_root(v_y^2 - 2ay)) / a</code></p> <p>we also have another equation, because at the apex the vertical velocity is 0:</p> <p><code>0 = v_y + at</code></p> <p>substitute:</p> <p><code>0 = v_y + (-v_y +/- square_root(v_y^2 - 2ay))</code></p> <p><code>0 = square_root(v_y^2 - 2ay)</code></p> <p><code>0 = v_y^2 - 2ay</code></p> <p><code>v_y = square_root(-2ay)</code>, or</p> <p><code>v_y = square_root(2gy)</code></p> <p>For v_x:</p> <p><code>v_x*t = x</code></p> <p>from before, t = v_y / a, so</p> <p><code>v_x = (x*g)/v_y</code></p> <p>I hope that made enough sense.</p>
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