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2 years ago

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An object is thrown off a 256-foot-tall building, and the distance of the object from the ground is measured every second. The f

unction that models the height, h, of the object after t seconds is h(t) = –16t2 + 96t + 256. Determine the time when the object hits the ground. After how many seconds does the object hit the ground? A.2 B.4 C. 8 D. 16

2 answers:

snow_tiger [21]2 years ago

7 0

Answer:

C. 8

Step-by-step explanation:

When the object hits the ground, h(t) = –16t^2 + 96*t + 256 = 0. Using the quadratic formula (where a= -16, b=96 and c=256) we get:

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

$t=\frac{-96 \pm \sqrt{96^2 - 4(-16)(256)}}{2(-16)}$

$t=\frac{-96 \pm 160}{-32}$

$t_1=\frac{-96 + 160}{-32}$

$t_1=-2$

$t_2=\frac{-96 - 160}{-32}$

$t_2=8$

The first root (-2) has no physical sense and is discarded. Then, the object hits the ground after 8 seconds.

hichkok12 [17]2 years ago

6 0

At 8 seconds the object will hit the ground ... Which is <span>C. 8

Hope This Is Sufficient !!</span>

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Step-by-step explanation:

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