Sam and Sarah have taken their math textbooks to the top of a 12 story building and look at the pool which is 160 straight below them. Sam..?

For Sam, The object hits the ground when its position is s(t) = 0. Set the position equation equal to zero and solve for t.

The position equation is s(t) = 16t^2 + v(o)t + h(o)
In Sam's case, v(o) = 0 and h(o) is 160ft.

Sam
16t^2 = 160 --->t = 3.2s
Sarah
16t^2+48t=160 --->t=2s

The pool is 160 below them?
Well...
160 miles or inches? I'm gonna assume that's 160 feet as that's about the corresponding height for a 12-story building.
I'll take acceleration of the Earth as 32.2 ft/s^2

Then use the displacement equation.
s = Vi*t + at^2 / 2
Using a as g, let's solve for Sam's book.
The initial V is 0, so the eq. is:
s = gt^2 / 2
160 ft = (32.2 ft/s^2)*t^2 / 2 = 3.15 s

Now Sarah (this will evolve into a quadratic equation):
160 ft = (48 ft/s)*t + (32.2 ft/s^2)*t^2 / 2
16.1t^2 + 48t - 160 = 0
t = ( -48 +-root(48^2 - 4*16.1*(-160)) ) / (2*16.1) = 2 s or -4.98 s.

After 2 s you hear splash #1 from Sarah, while it takes 3.15 s for Sam's book to get wet.

Best wishes!