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)=-0.01 x^{2}+0.7 x + 5.7 ), 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 17.95, which occurs 35 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?
(square) feet
(type an integer or decimal rounded to the nearest tenth as needed.)

Explanation:

Step1: Set height to zero

To find the maximum horizontal distance, we need to find the value of \( x \) when \( f(x) = 0 \). So we solve the quadratic equation \( -0.01x^{2}+0.7x + 5.7=0 \).

Step2: Use quadratic formula

For a quadratic equation \( ax^{2}+bx + c = 0 \), the solutions are given by \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \). Here, \( a=- 0.01 \), \( b = 0.7 \) and \( c = 5.7 \).
First, calculate the discriminant \( D=b^{2}-4ac=(0.7)^{2}-4\times(-0.01)\times5.7=0.49 + 0.228=0.718 \)
Then, \( x=\frac{-0.7\pm\sqrt{0.718}}{2\times(-0.01)}=\frac{-0.7\pm0.8473}{- 0.02} \)
We have two solutions:
\( x_1=\frac{-0.7 + 0.8473}{-0.02}=\frac{0.1473}{-0.02}=- 7.365 \) (we discard this negative solution as distance can't be negative)
\( x_2=\frac{-0.7-0.8473}{-0.02}=\frac{-1.5473}{-0.02}=77.365\approx77.4 \)

Answer:

77.4