The Math

When it comes to cannons not only is it important to understand the science that allows the cannon to fire, but we must also understand the mathematics behind it in order to determine certain information about the movement a shot from the cannon.

To determine such things we would use the following formula:

-16t^2+vt+h

t=Time in Seconds=x

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

h=Initial height



 To better understand here is an example:

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 should look like this -16t^2+192t+32

I'm going to start off by determining how long the cannon ball remains in the air. This bit of data is the positive x intercept of the formula. To determine the x intercept we must use the Quadratic formula that looks like this:



n.p how to write in quadratic formula. 2010. Web. 29 January 2011.



Filling in what we know we can simplify it down and 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.

For our cannon's angle we used the formula for velocity (speed (ft/sec))cos(launch angle). However there was one problem with this... we had no idea how fast a tennis ball can cannon could shoot so we assumed five feet persecond. Now for launch angle we just chose three eperate angles based on our research: 55, 45 and 35 degree's. Putting these value's in the function we reciever the following values respectiavley; 3.214,3.536,4.096. Naturally we figured the highest velocity would be the best, however at 35degrees we realized that we would have to have the base almost obstructing the cannon so we went with the solid 45 degrees.