Question

A rocket is launched with an initial upward velocity of 320 feet per second from an initial height of 15 feet.
The function h (t) = -1672 + 320t + 15 models the height of the rocket, in feet, t seconds after it was
launched. For how many seconds is the height of the rocket greater than or equal to 1.039 feet?

Answer 1

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.