Free Fall (kinematics-what I do wrong?)

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Homework Help Overview

The discussion revolves around a kinematics problem involving a rock tossed upwards and its subsequent fall into a hole. Participants are exploring the calculation of the rock's impact speed as it hits the bottom of a 10-meter deep hole.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply the kinematic equation to find the final velocity of the rock but expresses confusion over the correct setup and calculations. Some participants question the initial conditions and the interpretation of the velocity and position terms used in the problem.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's approach and calculations. There is a mix of interpretations regarding the initial conditions and the application of kinematic equations, indicating a productive exploration of the problem.

Contextual Notes

Participants note potential misunderstandings in the problem setup, such as the distinction between velocity and position, as well as the correct interpretation of the initial velocity of the rock.

Kalie
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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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