Free Fall (kinematics-what I do wrong?)

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The discussion focuses on calculating the impact speed of a rock tossed straight up with an initial velocity of +20 m/s that falls into a 10 m deep hole. The correct kinematic equation to use is Vf^2 = Vi^2 + 2aΔs, where the total distance of the fall is 30.41 m. The initial misunderstanding stems from misinterpreting +20 m as a location rather than a velocity. The final velocity should be calculated as negative, indicating direction, resulting in an impact speed of -24.4 m/s. Clarifying these points is essential for accurate physics problem-solving.
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A rock is tossed straight up with a velocity of +20m When it returns, it falls into a hole 10m deep.
What is the rock's velocity as it hits the bottom of the hole?
So basically it is asking for the impact speed right?
So that would be
Vf^2=Vi^2 +2adelta(s)
Vf^2=0+2*9.8*30.41
Vf^2=24.41
It says it is wrong. I know 30.41 is the right amount of the total meters in the fall since I solve another part and got the answer right. So that means that the kinematic equation is wrong so which would it be?
 
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never mind it is -24.4 instead of positive...
 
You should be careful when you type questions ,
you had said "with a velocity of +20 m" . . .
but +20m is a location, not a velocity.

you also ended with vf^2 = 24 , rather than vf = 24 ...
 
I always find it easy to simplify stuff in physics, like if a car going at 50m/s is chansing a car at 30m/s, it's good to think of a car going towards a stationary object at 20m/s. In this case, ignoring wind resistance, the rock will reach the point you throw at the same speed (note I don't say velocity) as you threw it up with. So it's like a rock falling down at 10m hole where initial velocity is 20m/s (if that is what you meant by 20m).
 
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