Question

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?

Answer 1

Answer:

$t = \frac{53}{32}$

Step-by-step explanation:

$h=\:-16^2+160t$

$9=\:-16^2+160t$      $substitute$

$-16^2+160t=9$

$-256+160t=9$

$-256+160t+256=9+256$

$160t=265$

$\frac{160t}{160}=\frac{265}{160}$

$t=\frac{53}{32}$

At  $\frac{53}{32}$  seconds it is at a height of 9 ft.