Question

a person shoots an arrow vertically into the air from a height of 6 feet with an initial velocity of 96 feet per second. the height, h, in feet above the ground, at any time, t (in seconds), is modeled by h(t) = 6 + 96t - 16t². practical range is the distance that the arrow starts from the ground (0 feet) to the minimum height of the arrow. c. the practical domain is the time the arrow starts (0 seconds) to the minimum height of the arrow. the practical range is the distance that the arrow starts from the ground (0 feet) to the time it takes to reach the ground. d. the practical domain is the time the arrow starts (0 seconds) to the time it takes to reach the ground. the practical range is the distance that the arrow starts from the ground (0 feet) to the maximum height of the arrow. e) use a graphing calculator to determine the horizontal intercepts. determine the practical meaning of these intercepts in this situation. select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. a. the horizontal intercepts are. the horizontal intercept has no meaning. the horizontal intercept indicates the time in seconds it takes the arrow to hit the ground. (type an ordered pair. use a comma to separate answers as needed. round to the nearest hundredth as needed.) b. the horizontal intercepts are. both have no meaning. (type an ordered pair. use a comma to separate answers as needed. round to the nearest hundredth as needed.) c. there are no horizontal intercepts.

Explanation:

Step1: Set up the equation

To find the horizontal intercepts, we set \(h(t) = 0\), so we have the equation \(0=6 + 96t-16t^{2}\). Rearranging this quadratic equation in standard form \(ax^{2}+bx + c = 0\), we get \(- 16t^{2}+96t + 6=0\), or multiplying both sides by - 1, \(16t^{2}-96t - 6=0\). Here, \(a = 16\), \(b=-96\), \(c=-6\).

Step2: Use the quadratic formula

The quadratic formula for a quadratic equation \(ax^{2}+bx + c = 0\) is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Substituting the values of \(a\), \(b\), and \(c\) into the formula:

First, calculate the discriminant \(\Delta=b^{2}-4ac=(-96)^{2}-4\times16\times(-6)=9216 + 384=9600\).

Then, \(t=\frac{96\pm\sqrt{9600}}{2\times16}=\frac{96\pm40\sqrt{6}}{32}=\frac{12\pm5\sqrt{6}}{4}\).

Calculating the two roots:

For the plus sign: \(t=\frac{12 + 5\sqrt{6}}{4}\approx\frac{12+5\times2.45}{4}=\frac{12 + 12.25}{4}=\frac{24.25}{4}\approx6.06\)

For the minus sign: \(t=\frac{12-5\sqrt{6}}{4}\approx\frac{12 - 12.25}{4}=\frac{-0.25}{4}\approx - 0.06\)

Step3: Analyze the practical meaning

Time cannot be negative in this context (the arrow is shot at \(t = 0\) and we are interested in the time after it is shot until it hits the ground). So the horizontal intercept \((-0.06,0)\) has no practical meaning, and the horizontal intercept \((6.06,0)\) represents the time (in seconds) when the arrow hits the ground.

Answer:

A. The horizontal intercepts are \((-0.06, 0)\), \((6.06, 0)\). The horizontal intercept \((-0.06, 0)\) has no meaning. The horizontal intercept \((6.06, 0)\) indicates the time in seconds it takes the arrow to hit the ground.