An object is dropped from 22 feet below the tip of the pinnacle atop a 1178-ft tall building. the height h of the object after t seconds is given by the equation h equals negative 16 t squared plus 1156. find how many seconds pass before the object reaches the ground.

t = 8.5 sec

Explanation:

Given,

Let 't' be the time taken by the object to reach the ground.

At the ground height = h = 0

From the expression of the height of the object in time 't' is

At the ground height of the ball is 0 m and the time taken is t

From the substitution of these values in the above equation, we get,

Hence, the required time taken by the ball to reach at the ground is 8.55 sec.

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~mia

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