Math Component

Cannon Blog: Math Component

Initial velocity equation: (speed (ft/sec))cos(launch angle)

Use the quadratic model h = -16t²+v₀t+h₀ to solve the following problem.

A cannonball is shot upward from the upper deck of a fort with an initial velocity of 192 feet per second.  The deck is 32 feet above the ground. 

Quadratic Model: h= -16t²+192t+32

highest point:

Since the parabola is symmetrical, the midpoint of the x in the parabola is the x in the vertex, and the y of the vertex will be the highest point of the parabola, which in this case, will result in the highest point of the cannon. 

The zeroes of this parabola is 0 and approximately 12.16sec

The midpoint of that will be (12.16+0)/2, which is 6.08(seconds).

-16(6.08)²+192(6.08)+32 = Highest point = 607.9

The highest point of the cannon is 607.9ft.

time:

-192±√(192)²-4(-16)(32)   =  12.16sec

                     2(-16)            

The cannon ball is in the air for approximately 12.16seconds.

1.     How high does the cannonball go?  607.9ft

2.     How long is the cannonball in the air? 12.16seconds

Or, you could've done it in the calculator ;)
It just looks more official to have the actual equation in there, in my opinion.

Justification for the degrees

In our cannon, we are using the 45 degree angle for it to result in the best possible launch. The General Ballistic Trajectory states that the 45 degree angle gives the maximum range of the launch.