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)=-4.9^2+268t+311.

Question

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)=-4.9^2+268t+311. For each of the answers below, put your numerical answer, rounded to 1 decimal place, in the first box and the units, spelled out completely, in the second box.

How long does it take until the rocket reaches its maximum height?  Correct Correct

What is the maximum height reached by the rocket?  Incorrect Correct

Assuming that the rocket will splash down into the ocean, how long does it take for splashdown to occur?   Incorrect 

Answer 1

The equation we need is h(t) = -4.9t² + 268t + 311

To find the time at which the rocket reaches its maximum height, we use the vertex form of a quadratic equation, which is h(t) = a(t−h)² + k, where the coordinates (h, k) are the vertex of the parabola.

Since h(t) = −4.9t²+ 268t + 311, a = -4.9, and the vertex formula is t = −b / 2a​ where b = 268, we can calculate the time (t) at which the rocket reaches its maximum height.

Therefore, t =− 268 / 2(−4.9) ≈ 27.4 seconds, the time it takes for the rocket to reach its maximum height.

To find the maximum height, we can substitute the time we found into the equation as h(t):

h(27.4) = −4.9(27.4)² + 268(27.4) + 311 ≈ 3,678.7 meters

That is the maximum height reached by the rocket.

To find the time for splashdown, we have to keep in mind the rocket has returned to sea level and so we need to find when h(t) = 0.

h(t) = −4.9t² + 268t + 311= 0

Solve this quadratic equation by using the quadratic formula:

t = ​​-b +/- √b² - 4ac / 2a, where a = -4.9, b = 268, and c = 311

t = −268 ± 2682 − 4(−4.9)(311) / 2(−4.9)

This will give two values of t, but we're interested in the positive one because time can't be negative.

t ≈ 68.7 seconds (this is the time needed for the rocket to splash down into the ocean)

Answer 2

h(t)=-4.9t^2+268t+311

take the derivative and set = 0

h' = -9.8t + 268=0

4.9t=134

t=about 27.347

t= about 27.3 seconds

that's the time to reach max height

then plug that t value into the original quadratic equation to solve for max height

h(134/4.9)=about 3975.5 meters

it hits the ocean when h(t)=0=-4.9t^2+268t+311

use the quadratic formula

t=268/9.8 +/- (1/9.8)sqr(268^2+4(4.9)(311))

t=about 55.8 seconds,

ignore the negative square root

55.8 is slightly more than twice the time to reach max height, 27.3 x 2 = 54.6

less than 1 second more.

given the 311 feet as relatively small compared to the max height of nearly 4000, that's what you'd expect

had it started at zero height instead of 311, then time to splashdown would have been

exactly twice the time to max height or exactly 54.6 seconds, rounded to nearest one decimal place