Question

test review #3, chapter 5
score: 3.5/37 answered: 3/37
question 4
nasa launches a rocket at ( t = 0 ) seconds. its height, in meters above sea - level, as a function of time is given by ( h(t)=-4.9t^{2}+175t + 164 ).
assuming that the rocket will splash down into the ocean, at what time does splashdown occur?
the rocket splashes down after (square) seconds.
how high above sea - level does the rocket get at its peak?
the rocket peaks at (square) meters above sea - level.
question help: video 1 video 2

Explanation:

Step1: Set height to 0 for splashdown

$$-4.9t^2 + 175t + 164 = 0$$

Step2: Apply quadratic formula $t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here $a=-4.9$, $b=175$, $c=164$:
$$t=\frac{-175\pm\sqrt{175^2-4(-4.9)(164)}}{2(-4.9)}$$

Step3: Calculate discriminant

$$\sqrt{30625 + 3214.4} = \sqrt{33839.4} \approx 183.95$$

Step4: Solve for positive t

$$t=\frac{-175 - 183.95}{-9.8} \approx \frac{-358.95}{-9.8} \approx 36.63$$

Step5: Find peak time (vertex of parabola)

$$t=-\frac{b}{2a} = -\frac{175}{2(-4.9)} \approx \frac{175}{9.8} = 17.86$$

Step6: Calculate peak height

$$h(17.86) = -4.9(17.86)^2 + 175(17.86) + 164$$
$$h(17.86) \approx -4.9(318.98) + 3125.5 + 164$$
$$h(17.86) \approx -1563.0 + 3125.5 + 164 \approx 1726.5$$

Answer:

The rocket splashes down after $\approx 36.63$ seconds.
The rocket peaks at $\approx 1726.5$ meters above sea-level.