Sure, let's go through the problem step by step.
### (a) How high is the arrow when it is first shot?
The height of the arrow when it is first shot can be found by evaluating the height equation at [tex]\( x = 0 \)[/tex]:
[tex]\[ y = -\frac{1}{25}(0)^2 + 0 + 3 \][/tex]
This simplifies to:
[tex]\[ y = 3 \text{ feet} \][/tex]
So, the height of the arrow when it is first shot is [tex]\( \boxed{3.00} \)[/tex] feet.
### (b) What is the maximum height of the arrow?
To find the maximum height of the arrow, we need to find the vertex of the parabola described by the quadratic equation [tex]\( y = -\frac{1}{25} x^2 + x + 3 \)[/tex]. The x-coordinate of the vertex of a parabola [tex]\( y = ax^2 + bx + c \)[/tex] is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = -\frac{1}{25} \)[/tex] and [tex]\( b = 1 \)[/tex]. Plugging in these values:
[tex]\[ x = -\frac{1}{2 \cdot -\frac{1}{25}} = -\frac{1}{-\frac{2}{25}} = \frac{25}{2} = 12.5 \][/tex]
Now, substitute [tex]\( x = 12.5 \)[/tex] back into the height equation to find the maximum height:
[tex]\[ y = -\frac{1}{25} (12.5)^2 + 12.5 + 3 \][/tex]
[tex]\[ y = -\frac{1}{25} \cdot 156.25 + 12.5 + 3 \][/tex]
[tex]\[ y = -6.25 + 12.5 + 3 \][/tex]
[tex]\[ y = 9.25 \text{ feet} \][/tex]
So, the maximum height of the arrow is [tex]\( \boxed{9.25} \)[/tex] feet.
### (c) How far horizontally does the arrow travel before hitting the ground?
The arrow hits the ground when the height [tex]\( y \)[/tex] is zero. We need to solve the quadratic equation for [tex]\( x \)[/tex] when [tex]\( y = 0 \)[/tex]:
[tex]\[ -\frac{1}{25} x^2 + x + 3 = 0 \][/tex]
This is a standard quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex]. To solve for [tex]\( x \)[/tex], we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = -\frac{1}{25} \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = 3 \)[/tex]. Plugging in these values:
[tex]\[ x = \frac{-1 \pm \sqrt{1^2 - 4\left(-\frac{1}{25}\right)(3)}}{2\left(-\frac{1}{25}\right)} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{1 + \frac{12}{25}}}{-\frac{2}{25}} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{\frac{25 + 12}{25}}}{-\frac{2}{25}} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{\frac{37}{25}}}{-\frac{2}{25}} \][/tex]
[tex]\[ x = \frac{-1 \pm \frac{\sqrt{37}}{5}}{-\frac{2}{25}} \][/tex]
[tex]\[ x = \frac{-1 \pm \frac{\sqrt{37}}{5}}{\frac{2}{-25}} \][/tex]
[tex]\[ x = \frac{-1 \pm \frac{\sqrt{37}}{5}}{-\frac{25}{2}} \][/tex]
[tex]\[ x = \frac{-1 \pm \frac{\sqrt{37}}{5}}{\frac{25}{-2}} \][/tex]
[tex]\[ x = \frac{-1 \pm \frac{\sqrt{37}}{5}}{\frac{25}{-2}} \][/tex]
Solving the expression under the square root and the division will lead to:
[tex]\[ x \approx 27.71 \text{ feet} \][/tex]
So, the horizontal distance before the arrow hits the ground is approximately [tex]\( \boxed{27.71} \)[/tex] feet.
By following these step-by-step calculations, we obtain the results for the initial height, maximum height, and horizontal distance traveled by the arrow.
