My dad and I worked together to find the velocity and angle for the most ideal launching. If you find an error or have a question, feel free to ask (and feel free to answer others’ questions because I might not know the answer) Post your discoveries and stuff if you want to.
There will be the most surface area if a ball enters the net perpendicularly. The net angle is approximately 36 degrees. We also approximated the distance to be 4 meters at 1 meter tall.
https://cdn.pbrd.co/images/1nGfmpEJ.png
The picture above shows a parabola with two known points. The standard form for a parabola is y = ax² + bx +c. The two points are plugged in.
(0,0) 0 = c
(4,1) 1 = 16a + 4b
Next, the slope needs to be calculated.
The derivative of standard form is dy/dx = 2ax + b.
https://cdn.pbrd.co/images/1nG8LPOY.png
The slope of the red line is
tan(270° + 36°) = -1.376
This is equal to the derivative at (4,1), so
2ax + b = -1.376
8a + b = -1.376
Now, we have two equations, so we can find a and b.
The elimination method results in
3.75276 = 2b
b = 1.876
1 = 16a + 4b
a = -0.4065
The equation for the parabola is
y = -0.4065x² + 1.876x
Next, we will find the maximum point of the parabola.
https://cdn.pbrd.co/images/1nGyTuyA.png
dy/dx = -0.813x + 1.876 = 0
0.813x = 1.876
x = 2.3075 thus x1 = 2.3075 and y1 = 2.1644
Using the acceleration formula,
y = Gt²/2
y1 = Gt1²/2
2.1644 = 9.8t1²1/2
t1 = 0.6646 (-.6646 can be removed because t cannot be negative)
Now that we know x1, y1, and t1, we can find Vx, which is constant throughout.
Vx = Change in x / Change in t
= x1/t1
= 3.472
The initial y velocity means the launch speed. It will be represented with V0y.
V0y = Gt1 because it will be the same as if it is dropped from the max value of the parabola.
V0y = Gt1 = 6.513
With Vx and V0y, the launch angle can be determined.
tan a = V0y/Vx
a = 61.938°
The initial velocity can be found, too.
V0 = √(Vx² + V0y²) = 7.3806 m/sec
Yay, we’ve made it this far. To summarize, 62° at 7.38 m/sec.
Bonus: Launch RPM Calculation
I need to state the conditions for this calculation. The wheel is 5", and it will be spinning at twice the velocity of the launch (n RPM).
2V0 = n * (1/60) * 2π * radius of wheel * 0.0254 (meter to inches)
n = (120 * V0)/(2πr) RPM
n = 2219.65 RPM
2219.65 RPM on a 5" wheel at 62°
I have made an Excel spreadsheet of gear ratios, so here is a picture of it.
https://cdn.pbrd.co/images/1nIBQ9GT.png