Initial velocity equation: (speed (ft/sec))cos(launch angle)
Use the quadratic model h = -16t2+v0t+h0 to solve the following problem.
Blog about the steps you took to solve these problems.
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 seconds
Procedure to find this answer
- In order to the this part of the project I plugged the formula into the y=. The formula is listed above and when i went over to the table I saw the highest part of the parabola, the x coordinate is 6 while the y coordinate is 608.
- Once I found this I used the number 608 as the highest that the cannonball goes because it is the vertext of the parabola.
- Afterwards I went back to the table and scrolled down until I reached zero or at least near it. I got to the x coordinate of 12 and the y coordinate of 32. If I went to 13 it would have been to the negatives so I assumed it was between 12-13 seconds of time.
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