MATH

-16t^2+vt+h

t=Time in Seconds=x

v= initial velocity: (speed (ft/sec))cos(launch angle)

h=Initial height



An example of the formula being used:

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. Determine the highest the cannonball goes and how long is it in the air? 

With this information our Equation looks like this -16t^2+192t+32

We start off by finding how long the cannon was in the air. This is found in the positive x-intercept. To determine the x-intercept we must use the Quadratic formula:



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We plug in the numbers to get the two x intercepts of: -.2 and 12.2. Since it is physically impossible for something to be in the air -.2 seconds our cannon shot was in the air for 12.2 seconds.

Now to determine how high the cannon shot goes. The formula x=-b/2a can be used to determine the axis of symmetry. Once we fill in our information (x=-192/2(-16)) we can solve to get 6=x. However since x=time this is not our answer. To answer our question we must plug 6 into the formula for t. With this our Formula should now look as such 16(6)^2+192(6)+32. Which after solving gives us our answer: The highest point the cannonball reaches is six seconds in adn it reaches about 608 feet off the ground.