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An object is launched straight up into
the air at an initial velocity of 32 feet
per second. It is launched from a height
of 3 feet off the ground. Its height h, in
feet, at t seconds is given by the
equation h = -8t^2 + 32t + 3. Find all
times t that the object is at a height of
27 feet off the ground.​

Respuesta :

sqdancefan

Answer:

  t ∈ {1, 3}

Step-by-step explanation:

You want to find t such that ...

  h = 27

  27 = -8t^2 +32t +3 . . . . . . substitute the expression for h

  24 = -8t^2 +32t . . . . . . . . . subtract 3

  -3 = t^2 -4t . . . . . . . . . . . . . divide by -8

  1 = t^2 -4t +4 = (t -2)^2 . . . . add 4 to complete the square

  ±√1 = t -2 . . . . . . . . . . . . . . take the square root

  t = 2 ± 1 . . . . . . . . . . . . . . . . add 2

  t = 1 or 3

The object is 27 ft off the ground at t = 1 and again at t = 3.

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