## Thursday, February 3, 2011

### Math Component

A cannnonball 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.

1. How high does the cannonball go?
2. How long is the cannonball in the air?

1.  To find out how high the cannonball goes, you need to find the vertex.
Formula: t= -b/(2a)
2.  If you substitute, and solve this--> t=(-192)/(2*-16)
3.  From there you should get 6... t= 6 seconds
4.  Then substitute 6 into the originial formula you had
h = -16t2+192t+32
h = -16(6)2+192(6)+32
h= 608 feet

To figure out number 2, you will need to use the quadratic model to figure out how long the cannonball will be in the air.
1.  -b ±√b2-4ac  You will need the quadratic formula to solve this problem
2(a)
2. Substitute your known data into the formula
-192 ±√1922-4(-16)(32)
2(-16)
3.  Once you solve this, you should get 12.17
4.  So the time that it is in the air would be
t= 12.16 seconds

1. How high does the cannonball go?                 608 feet
2. How long is the cannonball in the air?           12.16 seconds

#### 1 comment:

1. you did a great job on the math component Kelly :)
you listed the steps just right and organized your information/data properly! :D