To determine the time(s) $t$ when the rocket reaches a height of 370 feet, we can use the given height equation:

$h(t) = -16t^2 + 210t$

We want to find the value of $t$ when $h(t) = 370$ feet. So, we set up the equation:

$370 = -16t^2 + 210t$

This can be rearranged to:

$-16t^2 + 210t - 370 = 0$

This is a quadratic equation in the form $at^2 + bt + c = 0$, where:

• $a = -16$
• $b = 210$
• $c = -370$

We can solve this quadratic equation using the quadratic formula:

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

Substituting the values of $a$, $b$, and $c$:

$t = \frac{-210 \pm \sqrt{(210)^2 - 4(-16)(-370)}}{2(-16)}$ $t = \frac{-210 \pm \sqrt{44100 - 23680}}{-32}$ $t = \frac{-210 \pm \sqrt{20420}}{-32}$ $t = \frac{-210 \pm 142.9}{-32}$

This gives us two possible solutions:

1. $t_1 = \frac{-210 + 142.9}{-32} = \frac{-67.1}{-32} \approx 2.1$ seconds
2. $t_2 = \frac{-210 - 142.9}{-32} = \frac{-352.9}{-32} \approx 11.0$ seconds

So, the rocket will be 370 feet above the ground at approximately 2.1 seconds and 11.0 seconds after launch.

Would you like more details or have any questions? Here are some 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. At what time will the rocket return to the ground?
4. What is the significance of the coefficient $-16$ in the equation?
5. How would the time change if the height was 500 feet instead of 370 feet?

Tip: The quadratic formula is a reliable method for solving any quadratic equation, especially when it cannot be easily factored.