Math Component of Cannon

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.

Blog about the steps you took to solve these problems.

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 

1.     How high does the cannonball go? 607.9 feet

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

Calculations
-192√ (192)²-(4)(-16)(32)   =      12.16           
                   2(-16)
12.16/2= 6.08
-16(6.08)²+192(6.08)+32= 607.9 feet

Procedure to find this answer

•  In order to the this part of the project I plugged the formula into the y=. The formula is listed above and when i went over to the table I saw the highest part of the parabola, the x coordinate is 6 while the y coordinate is 608.
•  Once I found this I used the number 608 as the highest that the cannonball goes because it is the vertext of the parabola.
•  Afterwards I went back to the table and scrolled down until I reached zero or at least near it. I got to the x coordinate of 12 and the y coordinate of 32. If I went to 13 it would have been to the negatives so I assumed it was between 12-13 seconds of time.