Answer:
(a) 5 feet
(b) 19 feet
(c) 30.3 feet
To solve (a):
The initial height will be when x=0, because is in the kids hand. So we take x=0 in the formula:
[tex]y=-\frac{1}{14}(0)^2+2\cdot0+5=0+0+5=5[/tex]
THe height of the ball when leaves the childs hand, is 5 feet. THat's the answer for (a)
To solve (b):
To find how high the ball gets, we need to find the y coordinate of the vertex of the parabolla.
Then, using:
[tex]y-\text{vertex}=-\frac{b}{2a}[/tex]
Then
[tex]-\frac{2}{2\cdot(-\frac{1}{14})}=14[/tex]
x-coordinate of the vertex is 14. Now to find y-coordinate, we evaluate the function in x=14
[tex]y=-\frac{1}{14}(14)^2+2\cdot14+5=-14+28+5=19[/tex]
The answer to (b) is 19 feet.
To solve (c):
We need to know when the ball hit the ground. That means that y = 0, because y is the height. Then using the quadratic formula:
[tex]\begin{gathered} -\frac{2\pm\sqrt[]{2^2-4\cdot(-\frac{1}{14})\cdot5}}{2\cdot(\frac{-1}{14})}=14\pm\sqrt[]{266} \\ 14-\sqrt[]{266}=-2.3 \\ 14-\sqrt[]{266}=30.3 \end{gathered}[/tex]
One of the results is negative, so it doesn't have sense. The answer to (c) is 30.3 feet
