The equation for the rocket's height as a function of time is given by:

$h(t) = -4.9t^2 + 214t + 220$

To find the time at which the rocket splashes down, we need to solve for when the height $h(t) = 0$ (when it reaches sea level). This means solving the quadratic equation:

$-4.9t^2 + 214t + 220 = 0$

Solving this quadratic equation:

1. The quadratic equation is in the standard form $at^2 + bt + c = 0$, where:

   • $a = -4.9$
   • $b = 214$
   • $c = 220$

2. Use the quadratic formula:

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

Substituting the values:

$t = \frac{-214 \pm \sqrt{(214)^2 - 4(-4.9)(220)}}{2(-4.9)}$

$t = \frac{-214 \pm \sqrt{45796 + 4312}}{-9.8}$

$t = \frac{-214 \pm \sqrt{50108}}{-9.8}$

$t = \frac{-214 \pm 223.9}{-9.8}$

Now, calculate the two possible values for $t$:

$t_1 = \frac{-214 + 223.9}{-9.8} = \frac{9.9}{-9.8} = -1.01 \, \text{(not valid, as time cannot be negative)}$

$t_2 = \frac{-214 - 223.9}{-9.8} = \frac{-437.9}{-9.8} = 44.68$

So, the rocket splashes down after approximately 44.68 seconds.

Would you like a detailed explanation of any steps?

Related Questions:

1. What is the significance of the coefficient $-4.9$ in the equation for height?
2. How does the initial height $h(0) = 220$ influence the splashdown time?
3. Can we find the maximum height the rocket reaches? How?
4. What would happen to the splashdown time if the initial velocity was increased?
5. How can we interpret the two solutions from the quadratic equation in physical terms?

Tip:

Always discard any negative time solutions in these types of problems, as they are not physically meaningful.