SOLUTION: Can you please help my in How to solve quadratic function word problems
my question is
A ball is released 6 feet above the ground and thrown vertically into the air. The equation
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my question is
A ball is released 6 feet above the ground and thrown vertically into the air. The equation
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Question 604841: Can you please help my in How to solve quadratic function word problems
my question is
A ball is released 6 feet above the ground and thrown vertically into the air. The equation h=-16t^+112t+6 gives the height of the ball if the initial is 112 feet per second.
1. Write the equation of the axis of symmetry and find the coordinates of the vertex of the graph of the equation.
2. What is the maximum height above the ground that the ball reaches?
3. How many seconds is the ball in the air? Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! A ball is released 6 feet above the ground and thrown vertically into the air. The equation h=-16t^+112t+6 gives the height of the ball if the initial is 112 feet per second.
1. Write the equation of the axis of symmetry and find the coordinates of the vertex of the graph of the equation.
.
axis of symmetry:
x = -b/(2a)
x = -112/(2(-16))
x = -112/(-32)
x = 3.5 (vertical line crossing at x=3.5)
.
find, y-value when x=3.5:
-16t^+112t+6
-16(3.5)^+112(3.5)+6
202
So, vertex is at (3.5, 202)
.
2. What is the maximum height above the ground that the ball reaches?
202 feet
3. How many seconds is the ball in the air?
set h=0 and solve for t:
h=-16t^+112t+6
0=-16t^+112t+6
applying the "quadratic formula" gives us:
x {-0.1, 7.1}
we can throw out the negative solution (extraneous) leaving
x = 7.1 seconds
.
Details of quadratic formula follows:
