To determine how long it takes for the rocket to return to the ground, we need to find the time [tex]\( t \)[/tex] when the height [tex]\( h \)[/tex] of the rocket becomes zero. The height of the rocket is given by the quadratic equation:
[tex]\[ h = -5t^2 + 110t \][/tex]
We need to solve this equation for [tex]\( t \)[/tex] when [tex]\( h = 0 \)[/tex]:
[tex]\[ 0 = -5t^2 + 110t \][/tex]
This is a standard quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex]. In our equation:
- [tex]\( a = -5 \)[/tex]
- [tex]\( b = 110 \)[/tex]
- [tex]\( c = 0 \)[/tex]
To solve this quadratic equation, we can use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
First, we calculate the discriminant ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \Delta = (110)^2 - 4(-5)(0) \][/tex]
[tex]\[ \Delta = 12100 - 0 \][/tex]
[tex]\[ \Delta = 12100 \][/tex]
Now we substitute [tex]\(\Delta\)[/tex], [tex]\( a \)[/tex], and [tex]\( b \)[/tex] into the quadratic formula:
[tex]\[ t = \frac{-110 \pm \sqrt{12100}}{2(-5)} \][/tex]
Calculate the square root of the discriminant:
[tex]\[ \sqrt{12100} = 110 \][/tex]
Now we have:
[tex]\[ t = \frac{-110 \pm 110}{-10} \][/tex]
This gives us two solutions:
1. [tex]\( t_1 = \frac{-110 + 110}{-10} = \frac{0}{-10} = -0.0 \)[/tex]
2. [tex]\( t_2 = \frac{-110 - 110}{-10} = \frac{-220}{-10} = 22.0 \)[/tex]
Since negative time does not make sense in this context, we discard [tex]\( t_1 = -0.0 \)[/tex].
Therefore, the positive time value [tex]\( t_2 = 22.0 \)[/tex] seconds is the time it takes for the rocket to return to the ground.
Thus, the rocket returns to the ground approximately 22 seconds after launch.
