Using RT=D we can algebraically rearrange the equation to get R=D/T, or Terminal velocity equals Altitude divided by time. Since we have Altitude and drop time, from one measly data point, 350 ft and 4s, we can figure the rate, terminal velocity. Assuming that the flour bomb reaches terminal velocity almost immediately after being released, 350ft/3.28=106.7m. Everything here is metric. 106.7m/4s=26.68m/s. Now that we have the terminal velocity, we can figure the time for a flour bomb dropped at any altitude of equal mass, ~170g. To figure the drop time of a flour bomb at 200m altitude, we take the altitude y and divide it by terminal velocity Vt. T=y/Vt. With that we can then get the acceleration in the x direction. ax=Vt/T. ax=26.68m/s / 7.49s. ax=3.56m/s^2. Now that we have the x acceleration, we can use Pythagorean's Theorem to get the total acceleration, since we know the y acceleration. ax^2+ay^2=Atotal^2. Square root that and we get Atotal=10.44 m/s^2. Now we need the total velocity of the plane. VplaneTotal=vplane+vwind. Use a negative value for wind if the wind is a headwind, positive for a tailwind. With that, the acceleration, and the time we can then get the x displacement. x=V*T+.5*Atotal*T^2. We’ll set the plane’s speed at 25 m/s. x=25*7.49+.5*10.44*7.49^2. Solving that we get x=480.09m. Now that we have the x and y components of a triangle we can then get our angle. x=480.09, y=200, so, pythagoras again. 480.09^2+200^2=Side^2. Square root that and we get 520.08. Take that and divide the altitude by it and we get 0.385. Now acos(0.385), which equals 67.384º. That’s it for the math.

You do not have to create a GUI like I did, you can simply assign values to the various variables in the editor itself. For the code, simply input the equations and their variables in the order described above.

To use it, simply attach the scope to the hardware, input the variables into your simulator, get your angle, set it, and takeoff. When the target appears in the scope's crosshairs, release the flour bomb.