Question

A toy rocket launched into the air has a height (h feet) at any given time (t seconds) as

h = -16t2 + 160t until it hits the ground. At what time(s) is it at a height of 9 feet above the
ground?

Answer 1

Explanation:

Answer 2

t =0.06s and t=9.94 s

Explanation:

The equation that models the height of the rocket is

`h(t) = - 16 {t}^(2) + 160t`

when the rocket is at height of 9 fret above the ground, then

`h(t) = 9`

This implies that:

`- 16 {t}^(2) + 160t = 9`

Rewrite in standard quadratic form.

`- 16 {t}^(2) + 160t - 9 = 0`

The solution is given by

`t = \frac{ - b \pm \sqrt{ {b}^(2) - 4ac} }{2a}`

`t = \frac{ - 160 \pm \sqrt{ {160}^(2) - 4 * - 16 * - 9} }{2 * - 9}`

`t = 5 + (√(391) )/(4) \: or \: t = 5 - (√(391) )/(4)`

`t = 9.94 \: or \: .06`

The rocket reach the height of 9 feet at times t=0.06s and t=9.94 seconds