A fountain at a city park shoots a stream of water vertically from the ground. The function y=-8x^2+16 models the height of the stream of water in feet after x seconds.
A. What is the maximum height of the stream of water?
B. At what time does the stream of water reach its maximum height?
C. For how many seconds does the stream of water appear above ground?
This is a bogus equation. It describes a pet of water being shot horizontally, not vertically. It's missing an x term, which indicates the initial upward velocity.
B. As with all quadratics, the vertex is at x = -b/2a
A. See B above
C. solve for x in y=0
The equation for height above launch point in feet in time t is
y = Vi t - 16 t^2 where Vi is the initial speed up.
Either you have some strange typos or your math book does not do physics.
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To answer these questions, we need to analyze the given equation y = -8x^2 + 16. Let's break it down step by step:
A. What is the maximum height of the stream of water?
To find the maximum height of the stream of water, we need to determine the vertex of the parabolic function represented by the equation. The vertex of a parabola y = ax^2 + bx + c is given by the formula x = -b / (2a), and the maximum height y can be calculated by plugging this value of x back into the equation.
In our case, the equation is y = -8x^2 + 16, where a = -8, b = 0, and c = 16. Substituting these values into the vertex formula, we get:
x = -(0) / (2 * -8)
x = 0 / -16
x = 0
Now, substitute the value of x = 0 back into the equation y = -8x^2 + 16:
y = -8(0)^2 + 16
y = 0 + 16
y = 16
Therefore, the maximum height of the stream of water is 16 feet.
B. At what time does the stream of water reach its maximum height?
Since the equation represents the height of the stream of water after x seconds, the time at which the stream reaches its maximum height is equal to the value of x at the vertex. In this case, we determined in part A that x = 0 at the vertex. Hence, the stream reaches its maximum height at 0 seconds.
C. For how many seconds does the stream of water appear above ground?
To determine the duration for which the stream of water appears above ground, we need to find the time values when y (the height) is greater than zero. In other words, we need to solve the equation -8x^2 + 16 > 0.
Let's solve this inequality:
-8x^2 + 16 > 0
Divide both sides of the inequality by -8 (note we flip the direction of the inequality when dividing by a negative number):
x^2 - 2 < 0
Now, we need to find the values of x that satisfy this inequality. We can solve it by factoring or using the quadratic formula. Factoring is not possible in this case, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 1, b = 0, and c = -2. Substituting these values into the quadratic formula:
x = (0 ± √(0^2 - 4(1)(-2))) / (2 * 1)
x = (0 ± √(0 + 8)) / 2
x = (0 ± √8) / 2
x = ± √8 / 2
x = ± √4 * √2 / 2
x = ± 2√2 / 2
x = ± √2
Since we're only interested in the positive values of x, the stream of water appears above ground for x = √2 seconds.
So, the stream of water appears above ground for approximately 1.41 seconds.