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A ball is thrown from a height of 157 feet with an initial downward velocity of 9/fts . The ball's height h (in feet) after t seconds is given by the following. =h−157−9t-16tsquared How long after the ball is thrown does it hit the ground? Round your answer(s) to the nearest hundredth. (If there is more than one answer, use the "or" button.)

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sqdancefan

Answer:

  2.86 seconds

Step-by-step explanation:

We presume the height is measured above the ground, so the height when the ball hits the ground is 0 feet. Putting that into the equation and solving for t, we get ...

  0 = 157 -9t -16t^2

We can use the quadratic formula with a=-16, b=-9, c=157 to find the solution.

  t = (-b -√(b^2 -4ac))/(2a) . . . . . we are not interested in the negative solution

  t = (-(-9) -√((-9)^2 -4(-16)(157)))/(2(-16))

  t = (9 -√10129)/-32 = (9 -100.6429)/-32 ≈ 2.86384

The ball will hit the ground about 2.86 seconds after it is thrown.

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