2 Parabolas Intersecting at 2 Know Points.

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Finding the equations of two parabolas intersecting at known points (0,0) and (50, 3.44) requires defining both parabolas, but only one is provided. The tedious method of eliminating time (t) from the projectile motion equations yields the correct launch angles of approximately 31 and 63 degrees. An easier approach involves using software like VBA to automate calculations, reducing manual effort. Analytic geometry can help find the launch angle if the vertex and an additional point are known. Without sufficient information, solving for the angles directly from the given points remains complex.
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Is it possible to find the equations of 2 parabolas intersecting at 2 known points, (0,0) and (50, 3.44) for instance. Looking for quicker way to find angular range of projectile launched from (0,0) and passing through x=50m y=3.44m when launched at 25m/s. I got the right answer by the tedious (but pedagogically fruitful) method of eliminating (t) from X=VoCos@(t) and Y=VoSin@(t)-.5g(t^2) and then solving for the angle. I did it all symbolically resulting in a theta = arctan(a really messy quadratic). Plugged in the values and it spat out @={31,63}degrees, the correct answer.

Is there an easier way to solve this?
 
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Yes, you can find the intersecting points of two parabolas fairly easily but that is not going to solve your problem. If I understand your problem correctly, in order to solve for the intersecting points you actually need to have two defined parabolas, and you only have one.

If you don't like crunching numbers, try writing something in VBA or some other software. Then you only have to enter numbers and push a button.
 
I believe your solution is the easiest approach. You can find the launch angle using analytic geometry for parabolas but your problem doesn't give you enough information for that approach. Projectile motion can be described as a opening down parobola in the form (x-h)^2=-4a(y-k) where the vertex is (h,k) and the focus is at (h,k-a). If you were given the vertex and one additional point you would have the equation of the parabola. Taking the first derivative and evaluating at (0,0) would give you the slope of the tangent line at (0,0), the arctanget of which is the launch angle.
 
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