Certainly! To find all the values of [tex]\( t \)[/tex] for which the rocket's height is 30 meters, we need to solve the equation given by the height function for [tex]\( h = 30 \)[/tex] meters. The height function is:
[tex]\[ h = 50t - 5t^2 \][/tex]
We set this equal to 30 meters, giving us:
[tex]\[ 50t - 5t^2 = 30 \][/tex]
Next, we rearrange this equation to the standard quadratic form:
[tex]\[ -5t^2 + 50t - 30 = 0 \][/tex]
We can simplify this by dividing the whole equation by -5:
[tex]\[ t^2 - 10t + 6 = 0 \][/tex]
Now, we'll solve this quadratic equation. This can be done using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation [tex]\( t^2 - 10t + 6 = 0 \)[/tex], the coefficients are:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -10 \)[/tex]
- [tex]\( c = 6 \)[/tex]
Substitute these values into the quadratic formula:
[tex]\[ t = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(6)}}{2(1)} \][/tex]
[tex]\[ t = \frac{10 \pm \sqrt{100 - 24}}{2} \][/tex]
[tex]\[ t = \frac{10 \pm \sqrt{76}}{2} \][/tex]
[tex]\[ t = \frac{10 \pm 2\sqrt{19}}{2} \][/tex]
[tex]\[ t = 5 \pm \sqrt{19} \][/tex]
This gives us two solutions:
[tex]\[ t_1 = 5 + \sqrt{19} \][/tex]
[tex]\[ t_2 = 5 - \sqrt{19} \][/tex]
Calculating the approximate values:
[tex]\[ t_1 \approx 5 + 4.36 = 9.36 \][/tex]
[tex]\[ t_2 \approx 5 - 4.36 = 0.64 \][/tex]
Thus, the values of [tex]\( t \)[/tex] for which the rocket's height is 30 meters are:
[tex]\[ t \approx 0.64 \text{ seconds or } t \approx 9.36 \text{ seconds} \][/tex]
Therefore, rounded to the nearest hundredth:
[tex]\[ t = 0.64 \text{ seconds or } t = 9.36 \text{ seconds} \][/tex]
