Question

A projectile is launched in the air. The function h(t) = -16t^2 + 32t + 128 gives the height, h, in feet, of the projectile t seconds after it is launched. After how many seconds will the projectile land back on the ground?

Help is always greatly appreciated. :)

Answer 1

`\quad \huge \quad \quad \boxed{ \tt \:Answer }`

`\qquad \tt \rightarrow \:4 \:\: sec`

____________________________________

`\large \tt Solution \: :`

height of an object projected in air with time is represented by the given equation :

`\qquad\displaystyle \tt \rightarrow \: h(t) = - 16 {t}^(2) + 32t + 128`

The height when the object hits the ground will be 0

so, just equate the equation with 0, and we will get the value of t (time) when the projectile will be back to ground.

`\qquad\displaystyle \tt \rightarrow \: - 16 {t}^(2) + 32t + 128 = 0`

`\qquad\displaystyle \tt \rightarrow \: - 16( {t {}^(2) - 2t}^{} - 8) = 0`

`\qquad\displaystyle \tt \rightarrow \: t {}^(2) - 2t - 8 = 0`

`\qquad\displaystyle \tt \rightarrow \: t {}^(2) - 4t +2 t - 8 = 0`

`\qquad\displaystyle \tt \rightarrow \: t(t - 4) + 2(t - 4) = 0`

`\qquad\displaystyle \tt \rightarrow \: (t - 4)(t + 2) = 0`

So,

`\qquad\displaystyle \tt \rightarrow \: t = 4 \: \: \: \: or \: \: \: \: t = - 2`

but since time can't be negative, t = 4 seconds

The object will be back on ground after 4 seconds.

Answer 2

4 seconds

===========================

Given is the

quadratic function

, which has a graph of parabola.

It has two

x-intercepts

, with the greater one representing the

projectile on the ground, at h(t) = 0

.

Let's

solve

the equation:

- 16t² + 32t + 128 = 0

t² - 2t - 8 = 0

t² - 2t + 1 - 9 = 0

(t - 1)² = 9

t - 1 = 3 and t - 1 = - 3

t = 4

and t = - 2

The root with the greater value is

t = 4

, which represents the time we need.