Damon thank you for the previous help.
A geyser sends a blast of boiling water high into the air. During the eruption, the height (h) (in feet) of the water (t)seconds after being forced out from the ground could be modeled by h = -16t*t + 70t.
The last two questions are: how do you find the maximum height of the boiling water and how long is the boiling water in the air ? (using the quadratic formula)
h = -16 t^2 + 70 t
The maximum height can be found by calculus, by graphing h vs. t or "completing the square". They probably want you to complete the square. That means changing the equation to:
h = -16t^2 + 70t - 35 + 35
= -16(t^2 -70t/16 + 35^2/16^2)+ 35^2/16
= -16(t - 35/16)^2 + 35^2/16
The highest value of h occurs when t = 35/16, in which case the squared term becomes zero. The maximum height then is 35^2/16 = 76.6 feet
The water is in the air until h = 0 a second time. If you use the quadratic formula (with a = -16, b = 70 and c = 0), the answer is written
t = [-70 - sqrt(70^2]/-32 = 140/32 = 4.375 s
To find the maximum height of the boiling water, you need to determine the vertex of the quadratic equation. The vertex corresponds to the highest point on the graph, which represents the maximum height.
The quadratic equation that models the height of the water is given by h = -16t^2 + 70t. This equation is in the form of y = ax^2 + bx + c, where a = -16, b = 70, and c = 0.
To find the x-coordinate of the vertex, you can use the formula x = -b / (2a). In this case, x = -70 / (2*(-16)) = -70 / (-32) = 2.1875.
Now, substitute this value of t = 2.1875 into the equation h = -16t^2 + 70t to find the maximum height. h = -16(2.1875)^2 + 70(2.1875) ≈ 76.5625.
Therefore, the maximum height of the boiling water is approximately 76.5625 feet.
To determine how long the boiling water remains in the air, you need to find the time it takes for the height to reach zero, which corresponds to when the water touches the ground again.
Since the height is given by the quadratic equation h = -16t^2 + 70t, you need to solve this equation when h = 0.
Set the equation equal to zero: -16t^2 + 70t = 0.
Factor out t to get t(-16t + 70) = 0.
This equation is satisfied when t = 0 (which corresponds to when the water is initially forced out), and when -16t + 70 = 0.
Solving -16t + 70 = 0:
-16t = -70,
t = -70 / -16,
t = 4.375.
Therefore, the boiling water is in the air for approximately 4.375 seconds.