SOLUTION: Question 35679: A ball is thrown into the air with the velocity of 48 feet per second. With h for height and t for time and height determined with h=16tsquared+48t+4. How ma

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Question 35679: A ball is thrown into the air with the velocity of 48 feet per second. With h for height and t for time and height determined with h=16tsquared+48t+4. How ma      Log On


   

Question 35688: Question 35679:
A ball is thrown into the air with the velocity of 48 feet per second.
With h for height and t for time and height determined with h=16tsquared+48t+4.
How may seconds will the ball reach maximum height.
What is the maximum height.
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
I am assuming that you mean h=+-16t%5E2+%2B+48t+%2B+4. You can do this either algebraically or graphically. To do it algebraically, you need to realize that the graph is a paraola that opens down. The vertex of the parabola y=ax%5E2+%2Bbx+%2Bc always occurs at x=+-b%2F%282a%29.

In this problem a=-16, b=48, and c= 4, the vertex occurs at t=-b%2F%282a%29+=+-48%2F%282%2A%28-16%29%29=3%2F2 seconds.

To find the maximum height, substitute x=1.5 into the formula
h=+-16t%5E2+%2B+48t+%2B+4
h=+-16%2A%281.5%29%5E2+%2B+48%2A%281.5%29+%2B+4 (Got a calculator???
h+=+40+feet

Graphically, it is even easier. Just draw the graph y=+-16x%5E2+%2B+48x+%2B+4, and find the vertex by calculator methods. The x coordinate of the vertex is the time it takes to get there, and the y coordinate of the vertex is the highest point.
+graph%28+300%2C300%2C+-2%2C+5%2C+-4%2C+48%2C+-16x%5E2+%2B+48x+%2B+4%29

R^2 at SCC