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An object is dropped from the top of a tall building. The height S of the object after t seconds is given by the equation s=-16t^2+676. Find how many seconds pass before the object hits the ground.​ (Hint: The object is on the ground when S​ = 0.)

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sqdancefan

Answer:

  6.5 seconds

Step-by-step explanation:

When s=0, we have ...

  0 = -16t^2 +676

  t^2 = 42.25 . . . . . add 16t^2, divide by 16

  t = √42.25 = 6.5

6.5 seconds will pass before the object hits the ground.

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