Question

NASA launches a rocket at t = 0 seconds. Its height, in meters above sea level, as a function of time is

Answer 1

The time at which the splashdown would occur assuming the rocket will splash down into the ocean is 50.0 seconds.

Given the following data:

• Initial time = 0 seconds

• $h(t) = - 4.9t^2 + 238t + 353$

To calculate the time at which the splashdown would occur assuming that the rocket will splash down into the ocean:

The function of time.

Mathematically, the height of the rocket launched by NASA with respect to time is given by this function:

$h(t) = - 4.9t^2 + 238t + 353$

Note: The height is equal to zero (0) when the splashdown occur.

$h(t) = - 4.9t^2 + 238t + 353=0$

The formula for quadratic equation.

We would solve the above quadratic equation by using the quadratic formula:

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

Where:

• a = -4.9

• b = 238

• c = 353

Substituting the given parameters into the formula, we have;

$t = \frac{-238\; \pm \;\sqrt{238^2\; - \;4(-4.9)(353)}}{2(-4.9)}\\\\t = \frac{-238\; \pm \;\sqrt{56644\; + \;6,918.8}}{-9.8}\\\\t = \frac{-238\; \pm \;\sqrt{63562.8}}{-9.8}\\\\t = \frac{-238\; \pm \;252.12}{-9.8}\\\\t = \frac{-490.12}{-9.8}$

Time, t = 50.0 seconds.

Read more on time here: brainly.com/question/24728358