To find the time it takes for the spurt of water to return to the initial height, we need to determine when the height \( S \) of the water is equal to 0. The given quadratic equation representing the height of the water is:
[tex]\[ S = -16t^2 + 482 \][/tex]
We need to solve the equation \(-16t^2 + 482 = 0\) for \( t \).
1. Set the height equation equal to zero:
[tex]\[ -16t^2 + 482 = 0 \][/tex]
2. Rearrange the equation to isolate the quadratic term:
[tex]\[ -16t^2 + 482 = 0 \][/tex]
[tex]\[ -16t^2 = -482 \][/tex]
[tex]\[ t^2 = \frac{482}{16} \][/tex]
3. Simplify the right-hand side:
[tex]\[ t^2 = \frac{482}{16} \][/tex]
[tex]\[ t^2 = \frac{241}{8} \][/tex]
4. Take the square root of both sides to solve for \( t \):
[tex]\[ t = \pm \sqrt{\frac{241}{8}} \][/tex]
5. Simplify the square root expression:
[tex]\[ t = \pm \frac{\sqrt{241}}{2\sqrt{2}} \][/tex]
[tex]\[ t = \pm \frac{\sqrt{241}}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} \][/tex]
[tex]\[ t = \pm \frac{\sqrt{241 \cdot 2}}{2 \cdot 2} \][/tex]
[tex]\[ t = \pm \frac{\sqrt{482}}{4} \][/tex]
Hence, the solutions are:
[tex]\[ t = \pm \frac{\sqrt{482}}{4} \][/tex]
This means there are two times when the height is zero. One time is negative, which we can disregard because time cannot be negative in this context. Therefore, we consider the positive value:
[tex]\[ t = \frac{\sqrt{482}}{4} \][/tex]
Thus, the time it takes for the spurt of water to return to the jet's height is:
[tex]\[ t = \frac{\sqrt{482}}{4} \approx 3.88 \ \text{seconds} \][/tex]
So, the spurt of water returns [tex]\( 3.88 \)[/tex] seconds after the shot.
