Question

an athlete whose event is the shot - put releases a shot. when the shot whose path is shown by the graph to the right is released at an angle of 35°. its height, f(x), in feet, can be modeled by f(x)= - 0.01x² + 0.7x + 5.4, where x is the shots horizontal distance, in feet, from its point of release. use this model to solve parts (a) through (c) and verify your answers using the graph.
a. what is the maximum height of the shot and how far from its point of release does this occur? the maximum height is 17.6500 feet from the point of release. (type an integer or decimal rounded to four decimal places as needed.)
b. what is the shots maximum horizontal distance, to the nearest tenth of a foot, or the distance of the throw? (type an integer or decimal rounded to the nearest tenth as needed.)

Explanation:

Step1: Identify the function type

The height - distance function is a quadratic function \(f(x)=- 0.01x^{2}+0.7x + 5.4\), where \(a=-0.01\), \(b = 0.7\), \(c = 5.4\) and \(x\) is the horizontal distance.

Step2: Find the x - coordinate of the vertex

The x - coordinate of the vertex of a quadratic function \(y = ax^{2}+bx + c\) is given by \(x=-\frac{b}{2a}\). Substitute \(a=-0.01\) and \(b = 0.7\) into the formula:
\[x=-\frac{0.7}{2\times(-0.01)}=\frac{0.7}{0.02}=35\]

Step3: Find the maximum height

Substitute \(x = 35\) into the function \(f(x)=-0.01x^{2}+0.7x + 5.4\):
\[f(35)=-0.01\times(35)^{2}+0.7\times35 + 5.4\]
\[=-0.01\times1225+24.5 + 5.4\]
\[=-12.25+24.5 + 5.4\]
\[=17.65\]

Step4: For part b

The maximum horizontal distance occurs when the height \(y = 0\). So we set \(y=-0.01x^{2}+0.7x + 5.4=0\).
Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), with \(a=-0.01\), \(b = 0.7\), \(c = 5.4\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(0.7)^{2}-4\times(-0.01)\times5.4=0.49 + 0.216 = 0.706\).
Then \(x=\frac{-0.7\pm\sqrt{0.706}}{2\times(-0.01)}=\frac{-0.7\pm0.84}{-0.02}\).
We take the positive root \(x=\frac{-0.7 + 0.84}{-0.02}\) (since distance cannot be negative). \(x=\frac{0.14}{-0.02}=- 7\) (rejected) or \(x=\frac{-0.7-0.84}{-0.02}=\frac{-1.54}{-0.02}=77\)

Answer:

a. The maximum height of the shot is 17.65 feet and it occurs 35.0000 feet from the point of release.
b. The shot's maximum horizontal distance is 77.0 feet.