Cannon Blog: Math Component
Initial velocity equation: (speed (ft/sec))cos(launch angle)
Use the quadratic model h = -16t2+v0t+h0 to solve the following problem.
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
highest point:
Since the parabola is symmetrical, the midpoint of the x in the parabola is the x in the vertex, and the y of the vertex will be the highest point of the parabola, which in this case, will result in the highest point of the cannon.
The zeroes of this parabola is 0 and approximately 12.16sec
The midpoint of that will be (12.16+0)/2, which is 6.08(seconds).
-16(6.08)2+192(6.08)+32 = Highest point = 607.9
-192±√(192)2-4(-16)(32) = 12.16sec
2(-16)
The cannon ball is in the air for approximately 12.16seconds.
1. How high does the cannonball go? 607.9ft
2. How long is the cannonball in the air? 12.16seconds
I think this is perfect, the math equation looks very official ;) awesome job
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