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A model rocket is launched with an initial velocity of 250 ft per second. The height h, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 250t.
How many seconds after the launch will the rocket be 600 ft above the ground? Round to the nearest hundredth of a second. (Enter your answers as a comma-separated list.)

Sagot :

sqdancefan

9514 1404 393

Answer:

  {2.96, 12.66}

Step-by-step explanation:

A graph shows the rocket will be 600 feet up after 2.96 seconds, and again at 12.66 seconds.

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You can solve the equation h = 600 to find the times.

  -16t^2 +250t = 600

  -16(t^2 -15.625t) = 600

  -16(t^2 -15.625t +7.8125²) = 600 -16(7.8125²)

  -16(t -7.8125)² = -376.5625

  t -7.8125 = √(376.5625/16) ≈ ±4.8513

  t = 7.8125 ± 4.8513 ≈ {2.9612, 12.6638}

The rocket will be 600 ft above the ground 2.96 and 12.66 seconds after launch.

View image sqdancefan
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