Freefall prob that's been bugging me.

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Homework Help Overview

The problem involves a stone thrown from a bridge that strikes the water 10.0 m below after 1.70 s, with the goal of determining the stone's initial velocity. The context is centered around kinematics and the effects of gravity on motion.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of kinematic equations, particularly questioning the signs used for displacement and gravity. There is an exploration of coordinate systems and how they affect the interpretation of the problem.

Discussion Status

The discussion has revealed various interpretations of the coordinate system and the signs associated with displacement and gravity. Some participants have provided insights into the implications of choosing different signs, while others have shared personal experiences with similar mistakes in calculations.

Contextual Notes

There is an emphasis on the importance of sign conventions in kinematics, and participants note that time is always considered positive. The original poster expresses confusion regarding the relationship between initial velocity and displacement when the stone is thrown upwards.

Gersty
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This problem has been bugging me for a while...

If a stone thrown from a bridge strikes the water 10.0 m below after 1.70 s, what was the stone's initial velocity?My first attempt was to use Δy = vyi(Δt) +0.5(g) (Δt)^2 because I understand that the time to fall is dependent only on gravity. I keep getting an initial velocity of -14.22 m/s. The given answer is 2.45 m/s upward. What gives?
I get the feeling that I'm missing something simple.
 
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In your chosen coordinate system, is Δy a positive or negative value? How about g?
 
note that y is down, and gravity is pointing down. For positive t, your displacement is positive. if you subsitute +10, you will get -2.45 m/s. if you substitute -10, you will get -14.22 m/s. In the latter case, you are throwing the stone 10 m up in the air.
 
I used -10 for displacement, -9.81 for g. I was taught that time will always be positive.
"if you substitute -10, you will get -14.22 m/s. In the latter case, you are throwing the stone 10 m up in the air"...confuses me because if you are throwing the rock up in the air, it seems that the initial velocity and the displacement will both be positive, thus "up in the air".
 
Gersty said:
I used -10 for displacement, -9.81 for g. I was taught that time will always be positive.
This means that you choose the positive direction to be "up".
If you substitute the above values in the equation you will find v=+2.45m/s.
So is positive which means "up". This makes sense. If you just drop the stone (with no initial velocity), in 1.7 seconds it will fall about 14 m. Which is more than the 10m from the problem. So you need to delay it a little by throwing it upwards. It will go up for a while and then come down and end at 10m below the initial position after 1.7s.
 
Thanks so much for the help. After working through the problem like 6 times I finally figured out that I had all the concepts and values correct in terms of sign and direction. My mistake was in the algebra. I kept making the same little error with signs while doing the arithmetic. Guess I need more coffee.
 

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