Math Part:

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

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 parabola is symmetrical, so the midpoint of the x in the parabola is the x in the vertex. The y of the vertex is going to be the highest point of the parabola. This is the highest point in the cannons trajectory.  

The zeroes of this parabola is 0 and approximately 12.16sec

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

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

The highest point of the cannon is 607.9ft.

-192±√(192)²-4(-16)(32)   =  12.16sec

2(-16)

The cannon ball is in the air forapproximately12.16seconds.

1.     How high does the cannonball go?  607.9ft

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