SOLUTION
Step 1 :
In this question, we are given that the height ( in feet ) of a ball thrown by a child is :
[tex]\begin{gathered} y\text{ =}\frac{-1}{16}x^2\text{ + 2 x + 3 } \\ \text{where x is the horizontal distance in f}eet\text{ from the point at which } \\ the\text{ ball is thrown} \end{gathered}[/tex]Step 2 :
a) How high is the ball when it leaves the child's hand?
Answer:
[tex]\begin{gathered} \text{Given y = }\frac{-1}{16}x^2\text{ + 2 x + 3 } \\ \text{when it leaves the child's hand, x = 0} \\ y\text{ = }\frac{-1}{16}(0)^2\text{ + 2 ( 0 ) + 3 } \\ y\text{ = 3 f}eet \end{gathered}[/tex]b) What is the maximum height of the ball?
Answer:
[tex]\begin{gathered} \frac{d\text{ y}}{d\text{ x}}\text{ = }\frac{-1}{8}\text{ x + 2 = 0} \\ \frac{-1}{8}x\text{ = - 2} \\ \frac{x}{8}\text{ = 2 } \\ \text{x = 2 x 8} \\ x\text{ = 16 f}eet \\ \text{Put x = 16 into the equation:} \\ y\text{ = }\frac{-1}{16}x^2\text{ + 2 x + 3 } \\ y\text{ = }\frac{-1}{16}(16)^2\text{ + 2 ( 16 ) + 3 } \\ y\text{ = -16 + 32 + 3 } \\ y\text{ ( Ma}\xi mum\text{ Height ) = 19 fe}et \end{gathered}[/tex]c) How far from the child does the ball strike the ground?
[tex]\begin{gathered} \text{Given y = }\frac{-1}{16}x^2\text{ + 2 x + 3 } \\ \text{Set y = 0, we have that:} \\ \frac{-1}{16}x^2\text{ + 2 x + 3 = 0} \\ \text{Multiply both sides by 16, we have that:} \\ -x^2\text{ + 32 x + 48 = 0} \\ x^2\text{ - 32 x - 48 = 0} \\ \text{The values of x are: -1. 436 or 33. 436} \\ we\text{ ignore x = - 1. 436 ( since it is negative )} \end{gathered}[/tex]But the child threw the ball from x = 0,
It means that the ball landed 33. 436 feet away from the child.
