Homework question

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)=-4.9𝑡2+79𝑡+115.

Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?

The rocket splashes down after    seconds.

How high above sea-level does the rocket get at its peak?

The rocket peaks at    meters above sea-level.

Answer 1

h(t) = -4.9t^2 +79t +115 =0 solve for t=17.46616 seconds

to splash down (ignore -1.34371 the other x value making h=0)

max height is at time t = midpoint between 17.46616 and -1.34371, their average=8.061225

max height = h(8.061225) = -4.9(8.061225)^2 +79(8.061225)+115 meters

= about 433.42 meters

or you could take the derivative, set = 0 and solve for t= time at max height

h'(t) = -9.8t +79 = 0

t = 79/9.8 = 395/49 = 8.0612245 seconds to max height

h(395/49) = -4.9(395/49)^2 + 79(395/49) + 115 = max height = 42475/98 = about 433.41837 meters

Answer 2

Hi EJ, this is definitely a quadratic function and can be manipulated in a number of ways to find the answers. The first value we need to find is the time (in seconds) that the rocket splashes down, which is mathematically equivalent to finding the x-intercept. As a refresher, the x-intercept is the value of x (in this case--time) when y (meters) is equal to 0. We can achieve this either by using the quadratic formula or by completing the square. The final value we have to find is the number of meters above sea level that the rocket peaks at. This can be obtained by finding the vertex. As a reminder, the vertex is either the minimum point on an upward facing parabola or the maximum point on a downward facing parabola. We know, just by the question, but also since the leading coefficient a is negative, that we will be calculating the maximum point-- the peak value of the rocket.

We already know the y-intercept because we're given that 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) =-4.9t²+79t+115.

So if we plug in 0 for t, we get:

-4.9(0)² + 79(0) +115 = 115

Which proves that when time is 0, the rocket is at 115m

To start, let's find the vertex. The formula for finding the x value of the vertex is:

-b/2a

which yields:

-79/-4.9(2) --> -79/-9.8 --> 8.06

To find the y value, we just plug x into the original equation:

-4.9(8.06)² + 79(8.06) + 115 = -318.32 + 636.74 + 115 = 433.42

So, in 8.06 seconds (t), the rocket reaches its maximum point at 433.42 meters before it starts to descend.

The next item to find is the splash down value, or the time in seconds it takes for the rocket to have a y = 0. To do this, I am going to complete the square, although you could also use the quadratic formula; both works!

First, I will manipulate the expression to have the leading coefficient = 1. To do this, I will have to divide every value by -4.9:

t² - 16.22t - 23.469 = 0

Add 23.469 to both sides:

t² - 16.122t = 23.469

Now we attempt to complete the square.

(16.122/2)² = (8.061)² = 64.980

Now we can rewrite the expression to include the constant term that completes the square. Since we added

64.980 to the left side, we have to do the same to the right side to honor the equation:

t² - 16.122t + 64.980 = 23.469 + 64.980

(t - 8.061)² = 88.45

When you take the square root of each side, you get:

t - 8.061 = +- 9.405

*We'll only concern ourselves with the positive version since realistically, we wouldn't deal with negative seconds

Add 8.061 to both sides and you end up with 17.466. This means that when the time is 17.466 seconds, the rocket will splash into the ocean.

Answer 3

h(t) = -4.9t² + 79t + 155. Rocket is launched at a height of 115 meters. Height is = 0 when it splashes down.

to find splash-down time, set h(t) = 0

-4.9t² + 79t + 115 = 0.

May want to simplify first by dividing by -4.9

Get t² - 16.122t - 23.469.= 0

Can solve for t by quadratic formula, completing the square or on-line quadratic equation solver.

Splash-down occurs as t = 17.466 seconds.

The maximum height occurs at the vertex of the downward opening parabola. Calculate -b/2a, where b is 79 and a is -4.9. -79/2(-4.9) = -79/-9.8 = 8.061 seconds.to peak altitude.

Substituting 8.061 for t in the original equation yields a maximum height of 433.42 meters

I also suggest plotting the curve on Desmos Graphing Calculator or other method so you can visualize the trajectory.

Answer 4

The rocket splashes down when h(t) = 0.

Maximum height when t = -79 / [2(-4.9)] = 8.0612

Maximum height = h(8.0612)