Question

An object is dropped from 43 feet below the tip of the pinnacle atop a 367-ft tall building. The height h of the object after t seconds is given by the equation

h=-16t^2+324. Find how many seconds pass before the object reaches the ground.

Answer 1

An object is dropped from 43 feet below the tip of the pinnacle atop a 367 ft. tall building. Thus, its height from the ground, when it is dropped is 367-43 = 324 ft.

The height h of the object after t seconds is given by the equation h= -16t²+324 . Now, h = 0, when the object reaches the ground so that -16t²+324 = 0 or, 16t² =324 or, t² = 324/16 . Then t = √(324/16) = 18/4 4.5 seconds. Thus,4.5 seconds will pass before the object reaches the ground.

Note:

Since time cannot be negative, we have considered only the positive square root of (324/16).