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.

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.

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.

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 = 9.8*t1²*1/2

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