A toy rocket launched into the air has a height ( h feet) at given time ( t seconds) as h= -16^2+160t until it hits the ground. At what time is it at a height of 9 feet above the ground?

Respuesta :

Answer:

[tex]t = \frac{53}{32}[/tex]

Step-by-step explanation:

[tex]h=\:-16^2+160t[/tex]

[tex]9=\:-16^2+160t[/tex]      [tex]substitute[/tex]

[tex]-16^2+160t=9[/tex]

[tex]-256+160t=9[/tex]

[tex]-256+160t+256=9+256[/tex]

[tex]160t=265[/tex]

[tex]\frac{160t}{160}=\frac{265}{160}[/tex]

[tex]t=\frac{53}{32}[/tex]

At  [tex]\frac{53}{32}[/tex]  seconds it is at a height of 9 ft.