After the semester is over, you discover that the math department has changed textbooks (again) so the bookstore won't buy back your nearly-new book. You and your friend Anna decide to get creative. You go to the rook of a five-story building and look over the edge to the reflecting pool 60 feet below. You drop your book over the edge at the same instant that Anna chucks her book straight down at 30 feet per second. By about how many seconds does her book beat yours into the water (approximately)? Your function is s(t)=−16t2+60 Anna's function is s(t)=−16t2−30t+60

Since the position of the pool is at 60 feet below, s(t) = -60 for both me and Anna. So, both expressions for the motion of the book are equated to -60 and then solved to find the time they reach the water.

So, For me s(t)= −16t² + 60

-60 = −16t² + 60

collecting like terms, we have

-60 - 60 = −16t²

- 120 = −16t²

dividing both sides by - 16, we have

t² = -120/-16

t² = 7.5

t = √7.5

t = 2.74 s

For Anna, s(t)= −16t² - 30t + 60

-60 = −16t² - 30t + 60

adding 60 to both sides, we have

-60 + 60 = −16t² - 30t + 60 + 60

−16t² - 30t + 120 = 0

dividing through by -2, we have

8t² + 15t - 60 = 0

Using the quadratic formula to find t

[tex]t = \frac{-15 +/- \sqrt{15^{2} - 4 X 8 X -60} }{2 X 8} \\t = \frac{-15 +/- \sqrt{225 + 1920} }{16} \\t = \frac{-15 +/- \sqrt{2145} }{16} \\t = \frac{-15 +/- 46.31}{16} \\t = \frac{-15 - 46.31}{16} or t = \frac{-15 + 46.31}{16} \\t = \frac{-61.31}{16} or t = \frac{31.31}{16} \\t = -3.83 s or 1.96 s[/tex]

Since t' cannot be negative, t' = 1.96 s.

It takes Anna's book 1.96 s to reach the water, while it takes 2.74 s for my book to reach the water.

So, the time it takes Anna's book to beat mine is the difference between the time of reaching the water of each book. So Δt = t - t' = 2.74 s - 1.96 s = 0.78 s