Rock's Velocity at Bottom of Hole

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SUMMARY

The rock is tossed upwards with an initial velocity of 20 m/s and falls into a 10 m deep hole. The correct final velocity as it hits the bottom of the hole is 24 m/s, reported as positive in the textbook, despite the downward direction of motion. The confusion arises from the distinction between speed and velocity, where speed is a scalar quantity and velocity is a vector quantity that includes direction. The calculations using the equation v² = vo² + 2a(Y - Yo) confirm that the final speed is indeed 24 m/s, regardless of the direction of motion.

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Homework Statement



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?



Homework Equations


V = Vo + at
Y - Yo = volt + .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 = volt + .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.
 
Physics news on Phys.org
You are right. If the initial velocity is positive, the final one is negative. 24 m/s is the final speed.

ehild
 

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