The height in feet of a baseball after it is hit can be modeled by the function f(x) =-16x^2 + 100x, where x is the time in seconds after the ball is hit. Find the ball's maximum height and the time it takes the ball to reach this height. Then find how long the ball is in the air.
what you need is the vertex.
for any parabola of the form
y = ax^2 + bx + c
the x of the vertex is -b/(2a)
once you have the x of the vertex, plug into the original to get y
y will be your maximum, and it happens at the x value.
(make sure you get a max height of 156.25 ft )
To find the maximum height of the ball, we need to find the vertex of the quadratic function f(x) = -16x^2 + 100x.
The formula to find the x-coordinate of the vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by x = -b/2a.
In this case, a = -16 and b = 100. Plugging these values into the formula, we get:
x = -100 / (2 * -16)
x = -100 / -32
x ≈ 3.125
So, the ball reaches its maximum height at approximately 3.125 seconds.
To find the maximum height, we substitute this value of x back into the function:
f(3.125) = -16(3.125)^2 + 100(3.125)
f(3.125) = -16(9.765625) + 312.5
f(3.125) = -156.25 + 312.5
f(3.125) ≈ 156.25
Therefore, the ball reaches a maximum height of approximately 156.25 feet.
To find how long the ball is in the air, we need to determine when the ball hits the ground. Since the height of the ball is zero when it hits the ground, we can set f(x) = 0 and solve for x:
-16x^2 + 100x = 0
Dividing both sides by 4x (to simplify the equation):
-4x + 25 = 0
Adding 4x to both sides:
25 = 4x
Dividing both sides by 4:
x = 25/4
x = 6.25
So, the ball hits the ground after approximately 6.25 seconds.
To find the time the ball is in the air, we subtract the time it takes to reach the ground from the time it takes to reach the maximum height:
Time in air = (time to reach the ground) - (time to reach maximum height)
= 6.25 - 3.125
= 3.125 seconds
Therefore, the ball is in the air for approximately 3.125 seconds.
To find the ball's maximum height, we need to determine the vertex of the quadratic function f(x) = -16x^2 + 100x.
The x-coordinate of the vertex can be found using the formula: x = -b / (2a), where "a" and "b" are the coefficients of the quadratic equation. In this case, a = -16 and b = 100.
x = -100 / (2*(-16))
x = -100 / (-32)
x ≈ 3.125
So, the ball reaches its maximum height at approximately x = 3.125 seconds.
To find the corresponding y-coordinate (the height), substitute this value back into the function f(x) = -16x^2 + 100x:
f(3.125) = -16(3.125)^2 + 100(3.125)
f(3.125) ≈ 156.25 feet
Therefore, the ball's maximum height is approximately 156.25 feet.
To determine how long it takes for the ball to reach this height, we need to find the time when f(x) = 156.25.
-16x^2 + 100x = 156.25
Simplifying this equation:
16x^2 - 100x + 156.25 = 0
Now we can use the quadratic formula to find the time it takes for the ball to reach the maximum height:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
Here, a = 16, b = -100, and c = 156.25.
x = (-(-100) ± sqrt((-100)^2 - 4(16)(156.25))) / (2(16))
x = (100 ± sqrt(10000 - 10000)) / 32
x = 100 / 32
x ≈ 3.125
So, the ball takes approximately 3.125 seconds to reach its maximum height.
Finally, to find how long the ball is in the air, we need to find the time when the height is equal to zero (since the ball's flight ends when it lands). This means we need to solve the equation:
-16x^2 + 100x = 0
Factoring out the common term, we have:
x(-16x + 100) = 0
This equation is satisfied when either x = 0 or -16x + 100 = 0.
For x = 0, the ball is at the initial position (it hasn't been hit yet), so we discard this solution.
Solving -16x + 100 = 0:
-16x = -100
x = -100 / -16
x ≈ 6.25
Therefore, the ball is in the air for approximately 6.25 seconds.
h=-16x^2+100x=x(100-16x)
when is h zero? answer, when x= 100/16, and x=0
IT is a parabola, so the max apex is at 1/2 the time between the zeroes, or at time= 100/32
use that x to find max height in the original formula.