Question

An object is dropped from 24 feet below the tip of the pinnacle atop a 1468-ft tall building. The height h of the object after t seconds is given by the equatior h= - 16t2 + 1444. Find how many seconds pass before the object reaches the ground. seconds pass before the object reaches the ground. (Type an integer or a decimal.)​

Answer 1

9.5 seconds pass before the object reaches the ground.

Explanation:

Height of the ball:

The height of the ball after t seconds is given by the following equation:

`h(t) = -16t^2 + 1444`

Find how many seconds pass before the object reaches the ground.

This is t for which h(t) = 0. So

`h(t) = -16t^2 + 1444`

`-16t^2 + 1444 = 0`

`16t^2 = 1444`

`t^2 = (1444)/(16)`

`t^2 = 90.25`

`t = \pm √(90.25)`

Since it is time, we only take the positive value.

`t = 9.5`

9.5 seconds pass before the object reaches the ground.