Initial velocity equation: (speed (ft/sec))cos(launch angle)
Quadratic model h = -16t2+v0t+h0
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.
Quadratic Model: h=-16t2 +192t+32
1. How high does the cannonball go? __608_feet___
2. How long is the cannonball in the air? ___12.16 seconds__
My steps for completing this:
1. Put 192(ft/sec) for V0. Put 32 feet for the h0 in the equation.
2. Plug the formula above into a graphing calculator and than graph
3. Hit 2nd , graph to get to the table.Once you are their, find the vertex of the graph and that shows how high the cannonball goes.
4. To find how long the cannonball was in the air, you use the quadratic formula and find the positive x-intercept.
simple and straight to the point :)
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