A rock is tossed straight up with a speed of 20 m/s. When it returns, it falls into a hole 10 m deep.
What is the rock's velocity as it hits the bottom of the hole?
V = Vo + at
Y - Yo = Vot + .5at2
v2 = vo2 + 2a(Y - Yo)
Y - Yo = .5(Vo + V)t
The Attempt at a Solution
My answer is -24 m/s but the book reports positive 24 m/s. I don't see why this is the answer if the rock is moving downward. I've identified up as the positive direction. I got my answer by using equation v2 = vo2 + 2a(Y - Yo) where V was unknown, Vo =20 m/s, a=-9.8, displacement=-10. I know the calculator only gives positive square roots but you can have negative square roots also and since this object is moving down, shouldn't the velocity be -24 m/s. The question later asks for time and according to the book, the value is 4.5 seconds using equation Y - Yo = Vot + .5at2 but this only true if you plug is -24 m/s not simply positive 24 m/s. This is why I don't see why the answer book doesn't report -24 m/s as the final velocity before it hits the bottom of the whole. This is very frustrating.