SOLUTION: Greetings everyone! I could really use some help from one of the wise Math wizards out there! Could someone please help me solve this word problem? I am stumped: If a soccer

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Greetings everyone! I could really use some help from one of the wise Math wizards out there! Could someone please help me solve this word problem? I am stumped: If a soccer       Log On


   

Question 18328: Greetings everyone!
I could really use some help from one of the wise Math wizards out there! Could someone please help me solve this word problem? I am stumped:
If a soccer ball is kicked straight up with an initial velocity of 32 feet per second, then its height above the earth is a function of time given by s(t) = -16t^2 + 32t. Graph this function for "0" less than or equal to "t" less than or equal to 2. What is the maximum height reached by the ball? My head hurts already!
Thanks to all.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
To find the maximum height reached by the ball, you need to find the location of the vertex of the parabola which opens downward.
You know it opens downward because the x^2-coefficient is negative.
The x-coordinate of the vertex of a parabola is given by -b%2F2a when you quadratic equation is in standard form: ax%5E2+%2B+bx+%2B+c, and yours is already there.
-b%2F2a+=+-32%2F2%28-16%29 = 1
The x-coordinate of the vertex is x = 1
To find the y-coordinate, substitute this value of x (x=1) into the quadratic equation and solve for y.
y+=+-16%281%29%5E2+%2B+32%281%29
y+=+-16+%2B+32
y+=+16
The vertex is at (1, 16) which is a maximum for your parabola, so the maximum height reached by the ball is 16 feet.
For the graph, see below:
graph%28300%2C200%2C-5%2C5%2C-4%2C20%2C-16x%5E2%2B32x%29
P.S. For the headache, take two aspirin and call me in the morning!