Question

An object is dropped from 43 feet below the tip of the pinnacle atop a 527​-ft tall building. The height h of the object after t seconds is given by the equation h equals negative 16 t squared plus 484. Find how many seconds pass before the object reaches the ground.

Answer 1

5.5 sec

Explanation:

The height h of the object after t seconds is given by the equation

h =- 16 t²+ 484

The height would be zero when the object hits the ground. therefore equating the above equation to zero

0 = - 16 t²+ 484 (solving for t)

16 t²= 484

t² = 484/16

t²= 30.25

taking square root on B.S

t= 5.5 sec

Answer 2

5.5 seconds

Explanation:

The function that gives the height of the object after t seconds falling is:

h = -16t^2 + 484.

To find the time when the object reaches the ground, we just need to use the value of h = 0 in the equation, and then find the value of t:

0 = -16t^2 + 484.

16t^2 = 484

t^2 = 30.25

t = 5.5 seconds

So the object will reach the ground after 5.5 seconds