Math

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. 

 h = -16t²+v₀t+h₀

How high does the cannonball go?

How long is the cannonball in the air?

The height that the cannonball went is equal to the y value of the vertex

is the formula used to find the x value vertex which can then be plugged into the equation to find the y value or the height, b=initial velocity: 192 fps  a=-16

x= -192/(2)(-16)

x= -192/-32

x= 6

Before we plug x into the equation h = -16t²+v₀t+h₀ , we want to put the given variables in

h = -16t²+v₀t+h₀ 
h = -16t²+192t+32
h = -16(6)²+192(6)+32
h = -576 + 1152 + 32
h =  608

The maximum height of the cannonball is 608 feet


Now, we find how long the cannonball was in the air, ot the t value

is the formula used to find the t value, the a, b,

and c values are from the original formula  

The cannon ball was in the air for 12.16 seconds