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Solve (t - 3)2 = 6. The arrow is at a height of 48 ft after approximately _ s
and after _s​

User BradVoy
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2 Answers

1 vote

Answer: 0.55 & 5.45

Explanation:

User HPCS
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5 votes

Answer:

The arrow is at a height of 48 ft after approximately 0.55 s and after 5.45 s​

Explanation:

The following information is missing:

The height of an arrow shot upward can be given by the formula s = v0*t - 16*t², where v0 is the initial velocity and t is time.How long does it take for the arrow to reach a height of 48 ft if it has an initial velocity of 96 ft/s?

If the arrow is at a height of 48 ft and its initial velocity is 96 ft/s, then:

48 = 96*t - 16*t²

16*t² - 96*t + 48 = 0

16*(t² - 6*t + 3) = 0

t² - 6*t + 3 = 0

t² - 6*t + 3 + 6 = 0 + 6

t² - 6*t + 9 = 6

(t - 3)² = 6

t - 3 = √6

t - 3 = 2.45; t = 2.45 + 3; t = 5.45

or

t - 3 = -2.45; t = -2.45 + 3; t = 0.55

User Phil Figgins
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