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An object is dropped from 39 ft below the tip of the pinnacle atop of 363 ft tall building. The height h of the object after t seconds is h=-16t^2+324. How many seconds pass before reaching the ground

User Mklauber
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1 Answer

6 votes

Answer:

4.5 seconds pass before reaching the ground

Explanation:

The height of the object is given by the following equation:


h(t) = -16t^(2) + 324

How many seconds pass before reaching the ground

It hits the ground when h(t) = 0. So


h(t) = 0


-16t^(2) + 324 = 0


16t^(2) = 324


t^(2) = (324)/(16)


t = \pm \sqrt{(324)/(16)}


t = \pm 4.5

There are no negative instants of time, so only the positive answer interests to us.

4.5 seconds pass before reaching the ground

User Patilnitin
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