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

and after _s

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

Step-by-step 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