SOLUTION: Suppose Charlie O達rian of the Braves hits a baseball straight upward at 150 ft/sec from a height of 5 ft. a) Use the formula to determine how long it takes the ball to return t

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Suppose Charlie O達rian of the Braves hits a baseball straight upward at 150 ft/sec from a height of 5 ft. a) Use the formula to determine how long it takes the ball to return t      Log On


   

Question 268714: Suppose Charlie O達rian of the Braves hits a
baseball straight upward at 150 ft/sec from a height of 5 ft.
a) Use the formula to determine how long it takes the ball
to return to the earth.
b) Use the accompanying graph to estimate the maximum
height reached by the ball
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Suppose Charlie O達rian of the Braves hits a
baseball straight upward at 150 ft/sec from a height of 5 ft.
a) Use the formula to determine how long it takes the ball
to return to the earth.
h(t) = -16t^2 + 150t + 5 (I had to provide my own formula)
h = 0
-16t^2 + 150t + 5 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -16x%5E2%2B150x%2B5+=+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=%28150%29%5E2-4%2A-16%2A5=22820.

Discriminant d=22820 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-150%2B-sqrt%28+22820+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28150%29%2Bsqrt%28+22820+%29%29%2F2%5C-16+=+-0.0332156501954231
x%5B2%5D+=+%28-%28150%29-sqrt%28+22820+%29%29%2F2%5C-16+=+9.40821565019542

Quadratic expression -16x%5E2%2B150x%2B5 can be factored:
-16x%5E2%2B150x%2B5+=+%28x--0.0332156501954231%29%2A%28x-9.40821565019542%29
Again, the answer is: -0.0332156501954231, 9.40821565019542. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B150%2Ax%2B5+%29

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t =~ 9.408 seconds (Ignore the negative value)
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b) Use the accompanying graph to estimate the maximum
height reached by the ball
h(t) = -16t^2 + 150t + 5
I don't see a graph, so I'll use the equation
The max height is at the vertex, when t = -b/2a
t = -150/-32 = 75/16 seconds
h(75/16) =~ 356.5625 feet = max height