Thursday, February 3, 2011

Math Component

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 How high does the cannonball go? 608 feet How long is the cannonball in the air? 12.16 seconds 1. Plug 192 feet per second in for v0 and 32 feet for h0 2. Plug formula into calculator and graph 3. Go to the table and find the vertex of the graph (this is how high the cannonball went) 4. Using the quadratic formula, find the positive x-intercept (this is how long the cannonball was in the air) 5. Check your work
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3 comments:

  1. KnasserFebruary 3, 2011 at 6:10 PM

    I like how you gave a simple explanation of the steps that you used to find the answers.

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  2. zkhatibFebruary 3, 2011 at 7:50 PM

    I like your simple explanations. So easy a caveman could do it! I would just separate the numbered list to make it easier to read.

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  3. Sarah AnnetteFebruary 3, 2011 at 8:33 PM

    This is so great! I love how you labeled everything "Chemistry Component", etc. And the information is in your own words, not many people have done that. I agree with Zaid, I would align the numbers to make it easier to read, though.

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