To find the time $t$ when the rocket's height $h$ is 27 meters, we need to solve the equation:

$h = 60t - 5t^2$

Given that $h = 27$ meters, we substitute this into the equation:

$27 = 60t - 5t^2$

This is a quadratic equation, which can be rewritten as:

$5t^2 - 60t + 27 = 0$

To solve this quadratic equation, we'll use the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this equation, $a = 5$, $b = -60$, and $c = 27$. Substituting these values into the quadratic formula:

$t = \frac{-(-60) \pm \sqrt{(-60)^2 - 4(5)(27)}}{2(5)}$

Simplifying:

$t = \frac{60 \pm \sqrt{3600 - 540}}{10}$

$t = \frac{60 \pm \sqrt{3060}}{10}$

$t = \frac{60 \pm 55.36}{10}$

Now, we'll find the two possible solutions:

1. $t_1 = \frac{60 + 55.36}{10} = \frac{115.36}{10} = 11.54$
2. $t_2 = \frac{60 - 55.36}{10} = \frac{4.64}{10} = 0.46$

So, the two times at which the rocket's height is 27 meters are approximately:

$t = 0.46 \text{ seconds and } t = 11.54 \text{ seconds}$

Would you like more details or have any questions about this?

Here are 5 related questions:

1. What is the maximum height reached by the rocket?
2. How long does it take for the rocket to reach its maximum height?
3. What is the total time the rocket is in the air before it hits the ground?
4. What would be the height of the rocket after 2 seconds?
5. How would the height equation change if the initial velocity was different?

Tip: When solving quadratic equations, always check if both roots are physically meaningful in the context of the problem.