ANSWER :
a. y = 3 feet
b. The maximum height is 19 feet
c. 33.44 ft
EXPLANATION :
From the problem, we have the equation :
[tex]y=-\frac{1}{16}x^2+2x+3[/tex]
a. The height of the ball, y when x = 0 is :
[tex]\begin{gathered} y=-\frac{1}{16}(0)^2+2(0)+3 \\ \\ y=3 \end{gathered}[/tex]
b. The maximum height can be obtained by substitute x = h
where h is :
[tex]h=-\frac{b}{2a}[/tex]
Solve for h with a = -1/16 and b = 2
[tex]\begin{gathered} h=-\frac{2}{2(-\frac{1}{16})} \\ \\ h=16 \end{gathered}[/tex]
Substitute x = h = 16, then solve for y :
[tex]\begin{gathered} y=-\frac{1}{16}(16)^2+2(16)+3 \\ \\ y=19 \end{gathered}[/tex]
c. The distance of the ball to the child when the ball strikes the ground is when y = 0
[tex]\begin{gathered} y=-\frac{1}{16}x^2+2x+3 \\ \\ 0=-\frac{1}{16}x^2+2x+3 \end{gathered}[/tex]
Using quadratic formula with a = -1/16, b = 2 and c = 3
[tex]\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ x=\frac{-2\pm\sqrt{2^2-4(-\frac{1}{16})(3)}}{2(-\frac{1}{16})} \\ \\ x=\frac{-2\pm\sqrt{\frac{19}{4}}}{-\frac{1}{8}} \\ \\ x=\frac{-2\pm\frac{1}{2}\sqrt{19}}{-\frac{1}{8}} \\ \\ x=16\pm4\sqrt{19} \\ \\ \text{ In decimals,} \\ \\ x=16+4\sqrt{19}=33.44\text{ }ft \\ x=16-4\sqrt{19}=-1.44\text{ }ft \end{gathered}[/tex]
Since there's no negative distance, the answer will be 33.44 ft
