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


Steps for Solving:


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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4 comments:

  1. TravisFebruary 2, 2011 at 12:57 PM

    LDLDLDLDLD. Good job, Amanda! I got that too :)

    ReplyDelete
  2. AnonymousFebruary 3, 2011 at 8:20 AM

    Super nicee procedure, very detailed and easy to follow. But when I did it I didn't get 12.16 sec. how did you get that answer?

    ReplyDelete
  3. AmandaFebruary 3, 2011 at 8:32 AM

    You have to use the quadratic formula to solve for the x-intercept that represents the cannonball hitting the ground. When you look at teh table, it doesn't show you when it reaches the ground again. That's why you have to do it yourself.

    ReplyDelete
  4. AnonymousFebruary 3, 2011 at 2:42 PM

    I see now I was using 6 but thats the vertex. Thanks!

    ReplyDelete

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