Math Component

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.
Use this formula to solve the problem:  h = -16t²+v₀t+h₀

The quadratic model is:  -16t²+192t+32

1.     How high does the cannonball go? 608 feet

(Remember you are looking for a specific part of the vertex.)

2.     How long is the cannonball in the air? __12.15 seconds_

(Remember you can use the quadratic formula.)

How I solved:

     Mainly, I used my TI 83-Plus, which is a special graphing calculator. All I had to do was plug in the function into my "y=" key. I used "2nd TRACE" to find the maximum point of the line. That value shows how high the cannonball went in the air. (608 feet) I clicked the "TABLE" key to see how long the cannon was in the air, & looked for when the cannon was at 32 (the original point) again. The graph is a parabola, so it looks like a big upside-down"U." 

Take a look at the diagram I made for a clearer understanding: