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

`t = (53)/(32)`

Explanation:

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

`9=\:-16^2+160t`
`substitute`

`-16^2+160t=9`

`-256+160t=9`

`-256+160t+256=9+256`

`160t=265`

`(160t)/(160)=(265)/(160)`

`t=(53)/(32)`

At
`(53)/(32)` seconds it is at a height of 9 ft.