Question

an athlete whose event is the shot put releases the shot. when the shot is released at an angle of 35°, its path can be modeled by the formula ( y = - 0.01 x ^ { 2 } + 0.7 x + 5.9 ) in which x is the shots horizontal distance, in feet, and y is its height, in feet. this formula is shown by one of the graphs, (a) or (b), in the figure. use the formula to answer the questions below. use the formula to determine the shots maximum distance. the maximum distance is approximately (square) feet. (round to the nearest tenth as needed.)

Explanation:

Step1: Set \(y = 0\)

We want to find when the shot hits the ground (\(y = 0\)), so we set up the equation \(0=-0.01x^{2}+0.7x + 5.9\).

Step2: Use the quadratic formula

For a quadratic equation \(ax^{2}+bx + c=0\) (\(a=-0.01\), \(b = 0.7\), \(c = 5.9\)), the quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(0.7)^{2}-4\times(-0.01)\times5.9=0.49 + 0.236=0.726\).
Then \(x=\frac{-0.7\pm\sqrt{0.726}}{2\times(-0.01)}=\frac{-0.7\pm0.852}{-0.02}\).
We have two solutions:
\(x_1=\frac{-0.7 + 0.852}{-0.02}=\frac{0.152}{-0.02}=-7.6\) (discard since distance \(x\geq0\))
\(x_2=\frac{-0.7-0.852}{-0.02}=\frac{-1.552}{-0.02}=77.6\)

Answer:

\(77.6\)