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gogolik [260]
3 years ago
9

A model rocket is launched with an initial velocity of 240 ft/s. The height, h, in feet, of the rocket t seconds after the launc

h is given by
h = −16t2 + 240t.
How many seconds after launch will the rocket be 390 ft above the ground? Round to the nearest hundredth of a second.

s (smaller value)
s (larger value)
Mathematics
1 answer:
Setler [38]3 years ago
4 0

Answer:

About 1.85 seconds and 13.15 seconds.

Step-by-step explanation:

The height (in feet) of the rocket <em>t</em> seconds after launch is given by the equation:

h = -16t^2 + 240 t

And we want to determine how many seconds after launch will be rocket be 390 feet above the ground.

Thus, let <em>h</em> = 390 and solve for <em>t: </em>

<em />390 = -16t^2  +240t<em />

Isolate:

-16t^2 + 240 t - 390 = 0

Simplify:

8t^2 - 120t + 195 = 0

We can use the quadratic formula:

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

In this case, <em>a</em> = 8, <em>b</em> = -120, and <em>c</em> = 195. Hence:

\displaystyle t = \frac{-(-120)\pm \sqrt{(-120)^2 - 4(8)(195)}}{2(8)}

Evaluate:

\displaystyle t = \frac{120\pm\sqrt{8160}}{16}

Simplify:

\displaystyle t = \frac{120\pm4\sqrt{510}}{16} = \frac{30\pm\sqrt{510}}{4}

Thus, our two solutions are:

\displaystyle t = \frac{30+ \sqrt{510}}{4} \approx  13.15 \text{ or } t  = \frac{30-\sqrt{510}}{4} \approx 1.85

Hence, the rocket will be 390 feet above the ground after about 1.85 seconds and again after about 13.15 seconds.

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Given the data in the question;

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