Question

1. a rocket is launched in the air on an alien planet (different gravity than earth). data on the height of the rocket is found in this table:

(a) write an equation ( h(t) ) which gives the height of the rocket above the ground (in feet) after ( t ) seconds. (weve seen several equation writing strategies this quarter. be mindful of all the available options!)

(b) what is the maximum height the rocket will reach? how long will it take to reach that height?

(c) how long will it take for the rocket to hit the ground?

(d) calculate the aroc of ( h(t) ) on the interval ( 5,10 ). interpret your answer in the context of the problem.

Explanation:

Step1: Assume the equation form

Since the motion is vertical - motion under constant acceleration (on an alien planet), the general form of the height function is \(h(t)=at^{2}+bt + c\). When \(t = 0\), \(h(0)=0\), so substituting \(t = 0\) into \(h(t)=at^{2}+bt + c\) gives \(c = 0\). So \(h(t)=at^{2}+bt\).

Step2: Substitute values to find coefficients

When \(t = 1\), \(h(1)=a\times1^{2}+b\times1=a + b\), and \(h(1)=162\), so \(a + b=162\). When \(t = 2\), \(h(2)=a\times2^{2}+b\times2=4a + 2b\), and \(h(2)=288\), so \(4a + 2b=288\).
From \(a + b=162\), we have \(b=162 - a\). Substitute \(b = 162 - a\) into \(4a + 2b=288\):

$$
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$$

If \(a=-18\), then \(b=162-(-18)=180\). So \(h(t)=-18t^{2}+180t\).

Step3: Find the vertex of the parabola (for part b)

For a quadratic function \(y = ax^{2}+bx + c\) (in our case \(y = h(t)\), \(a=-18\), \(b = 180\)), the \(t\) - coordinate of the vertex is \(t=-\frac{b}{2a}\).

$$t=-\frac{180}{2\times(-18)} = 5$$

Substitute \(t = 5\) into \(h(t)=-18t^{2}+180t\):

$$h(5)=-18\times5^{2}+180\times5=-18\times25 + 900=-450+900 = 450$$

Step4: Find when the rocket hits the ground (for part c)

Set \(h(t)=0\), so \(-18t^{2}+180t=0\). Factor out \(-18t\): \(-18t(t - 10)=0\).
Using the zero - product property, \(t = 0\) (launch time) or \(t = 10\).

Step5: Calculate the AROC (for part d)

The average rate of change (AROC) of a function \(y = f(x)\) on the interval \([x_1,x_2]\) is \(\frac{f(x_2)-f(x_1)}{x_2 - x_1}\).
For \(h(t)=-18t^{2}+180t\), \(x_1 = 5\), \(x_2 = 10\)

$$h(5)=-18\times5^{2}+180\times5=450$$

$$h(10)=-18\times10^{2}+180\times10=-1800 + 1800=0$$

The AROC\(=\frac{h(10)-h(5)}{10 - 5}=\frac{0 - 450}{5}=-90\)

Answer:

(a) \(h(t)=-18t^{2}+180t\)
(b) The maximum height is \(450\) feet and it takes \(5\) seconds to reach that height.
(c) It takes \(10\) seconds for the rocket to hit the ground.
(d) The AROC of \(h(t)\) on the interval \([5,10]\) is \(-90\). This means that, on average, the height of the rocket is decreasing at a rate of \(90\) feet per second over the time interval from \(t = 5\) seconds to \(t = 10\) seconds.