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

Respuesta :

The Solution:

Given:

[tex]h=-16t^2+1444.[/tex]

We are required to find t when h = 0.

[tex]\begin{gathered} -16t^2+1444=0 \\ \\ -16t^2=-1444 \end{gathered}[/tex]

Divide both sides by -16.

[tex]t^2=\frac{-1444}{-16}=90.25[/tex][tex]\begin{gathered} t=\sqrt{90.25} \\ \\ t=9.5\text{ or }t=-9.5 \end{gathered}[/tex]

Thus, the correct answer is 9.5 seconds.