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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?


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Answer :

mhanifa

Answer:

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.

AǫᴜᴀWɪᴢ

[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]

[tex]\qquad \tt \rightarrow \:4 \:\: sec [/tex]

____________________________________

[tex] \large \tt Solution \: : [/tex]

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

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

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.

[tex]\qquad\displaystyle \tt \rightarrow \: - 16 {t}^{2} + 32t + 128 = 0[/tex]

[tex]\qquad\displaystyle \tt \rightarrow \: - 16( {t {}^{2} - 2t}^{} - 8) = 0[/tex]

[tex]\qquad\displaystyle \tt \rightarrow \: t {}^{2} - 2t - 8 = 0[/tex]

[tex]\qquad\displaystyle \tt \rightarrow \: t {}^{2} - 4t +2 t - 8 = 0[/tex]

[tex]\qquad\displaystyle \tt \rightarrow \: t(t - 4) + 2(t - 4) = 0[/tex]

[tex]\qquad\displaystyle \tt \rightarrow \: (t - 4)(t + 2) = 0[/tex]

So,

[tex]\qquad\displaystyle \tt \rightarrow \: t = 4 \: \: \: \: or \: \: \: \: t = - 2[/tex]

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

The object will be back on ground after 4 seconds.

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