Catapult
The skills we are exploring in this project are solving equations and making graphs. We will be launching a projectile determining the quadratic equation that models its movement, and graphing its trajectory in order to hit a target.
Our original plan was to build an air cannon out of PVC and compressed air, but we found that the air cannon would not be consistent in the trajectory so we changed the idea to a catapult because we figured it would be more consistent. We thought that the air-cannon would be inconsistent because we would have to pressurize the reservoir to the exact same pressure for every shot. Catapults are powered by spring, which are more consistent.
It gave me a better understanding on exactly how to map out a trajectory of a projectile. Also just how complicated figuring out the arch of a specific object with only three points as a reference. It gives me a better respect for the people who do this for a living.
A parabola is a graph of a quadratic equation that is used to determine the trajectory of an object. A quadratic equations is a second degree equation that you have to square to get an answer. For example, this is the way to find your parabola on your graph after putting it in. In other words, determining the height and distance an object goes, or the curve of an object such as a building structure. This is used in our project by determining the trajectory of our projectile. Parabolas also can be used for angles in architecture for buildings in domes and bridges. In both you need to know how a parabola works to make sure that the dome won’t collapse or the bridge won’t break. Other things parabolas are used for are curves on skis, it is used for accuracy and when you hit a bump in the snow, you don’t go flying off and hurt yourself.
We use intercept form because we had two “x” intercepts for our graph. They were the starting point (the catapult) and our ending point (projectile landing point).
The way we figured out our graph from this equation was entering the intercept points into “p” and “q” then put a vertex point into the “y” and “x” places in the equation. We then converted it into standard form by distributing properties. We then solved for “a”. After that we inserted random co-ordinance into our graph in order to find our curve.
In order to accomplish our task we have to launch our projectile without a target. We measure how far the projectile goes starting from the origin, aka the catapult. These two points, the origin and where the projectile lands, are the X-intercepts of our graph. Take the equation into a graph and once we have this we will be able to determine the exact distance. So if the target is 14” high, we will know by looking at our graph that it will be 166” away. To find the vertex we took the total distance fired, halved that and took that distance away from a wall, shot the wall, and measured the height of the hit on the wall. This gave us the vertex. We will determine where to place our catapult by looking at our graph and placing our catapult at its specified distance.
A real world application would be for military purposes. This would apply to the targeting systems of artillery. Basically the same thing we are doing but with explosives and greater distances.
This is 3-D graphing that we did in our comp books last semester.
