Question 536600: A person standing cloes to the edge on the top of a 200-foot building throws a baseball vertically upward. The quadratic function
s(t)=-16t^2+64t+200
models the ball's height above the ground, s(t), in feet, t seconds after it was thrown.
A)After how many seconds does the ball reach it's maximum height? What is the maximum height?
B)How many seconds does it take until the ball finally hits the ground?
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! A person standing cloes to the edge on the top of a 200-foot building throws a baseball vertically upward. The quadratic function
s(t)=-16t^2+64t+200
models the ball's height above the ground, s(t), in feet, t seconds after it was thrown.
A)After how many seconds does the ball reach it's [sic] maximum height? What is the maximum height?
The equation is a parabola, the max height is at the vertex.
The vertex is on the line of symmetry, which is
t = -b/2a
t = -64/-32 = 2 seconds
s(2) = -16*4 + 64*2 + 200
max ht = 264 ft
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B)How many seconds does it take until the ball finally hits the ground?
Find t when s(t) = 0
-16t^2 + 64t + 200 = 0
2t^2 - 8t - 25 = 0
| Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc) |
Quadratic equation  (in our case  ) has the following solutons:
For these solutions to exist, the discriminant  should not be a negative number.
First, we need to compute the discriminant :  .
Discriminant d=264 is greater than zero. That means that there are two solutions:  .
Quadratic expression  can be factored:
Again, the answer is: 6.06201920231798, -2.06201920231798.
Here's your graph:
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Ignore the negative solution
t =~ 6.062 seconds
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PS it's = it is
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