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Serjik [45]
3 years ago
14

A rocket is launched with an initial upward velocity of 320 feet per second from an initial height of 15 feet.

Mathematics
1 answer:
liberstina [14]3 years ago
5 0

Answer:

The height of the rocket is greater than or equal to 1.039 feet during first 19.936 seconds.

Step-by-step explanation:

The height of the rocket launched from the ground is modelled after the following expression:

h(t) = -16.087\cdot t^{2}+320\cdot t +15 (1)

Where:

t - Time, measured in seconds.

h - Height, measured in feet.

After a careful reading to the statement, we translate all relevant information into the following mathematical inequation:

-16.087\cdot t^{2}+320\cdot t +15\ge 1.039 (2)

After some algebra, we get the equivalent inequation:

-(16.087\cdot t^{2}-320\cdot t -13.961)\ge 0

16.087\cdot t^{2}-320\cdot t -13.961 \le 0

(x-19.936)\cdot (x+0.044)\le 0

Which means that the following conditions must be observed to satisfy the inequation above:

x-19.936\le 0\,\land\,x+0.044\ge 0

x\le 19.936\,s\,\land\,x\ge -0.044\,s

0\,s\le x \le 19.936\,s

The height of the rocket is greater than or equal to 1.039 feet during first 19.936 seconds.

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