Question

An object is dropped from 27 feet below the tip of the pinnacle atop a 1471-ft tall building. The height h of the object after t seconds is giveh= - 16t^2 + 1444. Find how many seconds pass before the object reaches the ground.How many seconds pass before the object reaches the ground

Answer 1

The Solution:

Given:

`h=-16t^2+1444.`

We are required to find t when h = 0.

`\begin{gathered} -16t^2+1444=0 \\ \\ -16t^2=-1444 \end{gathered}`

Divide both sides by -16.

`t^2=(-1444)/(-16)=90.25`
`\begin{gathered} t=√(90.25) \\ \\ t=9.5\text{ or }t=-9.5 \end{gathered}`

Thus, the correct answer is 9.5 seconds.