SOLUTION: The height h (in feet) above the ground of a baseball depends upon the time t (in seconds) it has been in flight. Cameron takes a mighty swing, but hits a bloop sinlge whose height

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: The height h (in feet) above the ground of a baseball depends upon the time t (in seconds) it has been in flight. Cameron takes a mighty swing, but hits a bloop sinlge whose height      Log On


   

Question 1013573: The height h (in feet) above the ground of a baseball depends upon the time t (in seconds) it has been in flight. Cameron takes a mighty swing, but hits a bloop sinlge whose height is described approximately by the equation h=-5t^2+45t+17. Ho long is the ball in the air? The ball reaches its maximum heght after how many seconds of light? What is the maximum height?
Found 2 solutions by Alan3354, lwsshak3:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The height h (in feet) above the ground of a baseball depends upon the time t (in seconds) it has been in flight. Cameron takes a mighty swing, but hits a bloop sinlge whose height is described approximately by the equation h=-5t^2+45t+17.
----
Ho long is the ball in the air?
The ball reaches its maximum height after how many seconds of light?
Max ht is the vertex of the parabola.
@t = -b/2a = -45/-10 = 4.5 seconds
----
How long is the ball in the air?
h=-5t^2+45t+17
Find t when h = 0
-5t^2+45t+17 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -5x%5E2%2B45x%2B17+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2845%29%5E2-4%2A-5%2A17=2365.

Discriminant d=2365 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-45%2B-sqrt%28+2365+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2845%29%2Bsqrt%28+2365+%29%29%2F2%5C-5+=+-0.363126566315131
x%5B2%5D+=+%28-%2845%29-sqrt%28+2365+%29%29%2F2%5C-5+=+9.36312656631513

Quadratic expression -5x%5E2%2B45x%2B17 can be factored:
-5x%5E2%2B45x%2B17+=+%28x--0.363126566315131%29%2A%28x-9.36312656631513%29
Again, the answer is: -0.363126566315131, 9.36312656631513. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-5%2Ax%5E2%2B45%2Ax%2B17+%29

================
Ignore the negative.
t =~ 9.363 seconds
-----
What is the maximum height?
= -5*4.5^2 + 45*4.5 + 17
= 118.25 feet
-----------------
Usually -5t^2 or -4.9t^2 is use for heights in meters.
-16t^2 is used for feet.

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
The height h (in feet) above the ground of a baseball depends upon the time t (in seconds) it has been in flight. Cameron takes a mighty swing, but hits a bloop sinlge whose height is described approximately by the equation h=-5t^2+45t+17. How long is the ball in the air? The ball reaches its maximum heght after how many seconds of light? What is the maximum height?
h=-5t^2+45t+17
complete the square
h=-5(t^2+9t+20.25)+101.25+17
h=-5(t+4.5)^2+118.25
This is an equation of a parabola that opens downward wwith vertex at (4.5,118.25)
maximum height of118.25 ft is reached after 4.5 seconds of flight.
Set h=0
=-5t^2+45t+17=0
solve for t by quadratic equation:
t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a=-5, b=45, c=17
ans: t=9.36
How long is the ball in the air? 9.36 seconds