An object is dropped from 26 feet below the tip of the pinnacle atop a 702 ft tall building. The height h of the object after t seconds is given by the equation h=-16t^2+676. Find how many seconds pass before the object reaches the ground.

Question

Physics

Elvis Patterson

15 September, 08:43

An object is dropped from 26 feet below the tip of the pinnacle atop a 702 ft tall building. The height h of the object after t seconds is given by the equation h=-16t^2+676. Find how many seconds pass before the object reaches the ground.

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Answers (1)

Know the Answer?

Pierce Bullock · Answer 1

6.5 seconds.

Explanation:

Given: h=-16t²+679

When the object reaches the ground, h=0.

∴ 0=-16t²+679

collecting like terms,

⇒ 16t²=679

Dividing both side of the equation by the coefficient of t² i. e 16

⇒ 16t²/16 = 679/16

⇒ t² = 42.25

taking the square root of both side of the equation.

⇒ √t² = √42.25

⇒ t = 6.5 seconds.