Does Instantaneous Velocity Account for Launch Height in Kinematics?

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SUMMARY

The discussion centers on the relationship between instantaneous velocity and launch height in kinematics, specifically in the context of objects falling under gravity. The formula y(t) = y0 + by - 0.5gt² is referenced, highlighting that while the initial height (y0) drops out when calculating instantaneous velocity, the time taken to reach a final height (yf) differs for objects launched from varying heights. Consequently, an object launched from a greater height will have a higher velocity upon reaching the same final height due to the longer time of descent.

PREREQUISITES
  • Understanding of kinematic equations, specifically y(t) = y0 + by - 0.5gt²
  • Basic knowledge of gravitational acceleration (g)
  • Familiarity with the concept of instantaneous velocity
  • Introduction to dimensionless analysis in physics
NEXT STEPS
  • Study the derivation and application of kinematic equations in various scenarios
  • Learn about the principles of gravitational motion and free fall
  • Explore the concept of instantaneous velocity in more complex systems
  • Investigate dimensionless analysis and its role in transitioning from Galilean to Newtonian physics
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and gravitational motion, as well as educators seeking to clarify concepts related to instantaneous velocity and launch height.

starstruck_
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<Moderator's note: Split from another thread and thus no template.>
I am working on my special relativity assignment right now, and it said to find the velocity of an object when it hits the ground. The height the object is launched at is > 0.

Question: comparing that velocity to the instantaneous velocity, does the instantaneous velocity not account for the fact that an object would have a higher velocity at a given point if it’s launched from a greater height?

Say you had two objects whose height can be modeled by the function y(t) = y0+ by-0.5gt^2

When you find the instantaneous velocity, the y0 term disappears. However, we know that the velocity of an object would be greater if it has a greater height.

What’s the difference here?
 
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starstruck_ said:
<Moderator's note: Split from another thread and thus no template.>
I am working on my special relativity assignment right now, and it said to find the velocity of an object when it hits the ground. The height the object is launched at is > 0.

Question: comparing that velocity to the instantaneous velocity, does the instantaneous velocity not account for the fact that an object would have a higher velocity at a given point if it’s launched from a greater height?

Say you had two objects whose height can be modeled by the function y(t) = y0+ by-0.5gt^2

When you find the instantaneous velocity, the y0 term disappears. However, we know that the velocity of an object would be greater if it has a greater height.

What’s the difference here?
The question is a bit confused. If I understand correctly, you are considering two objects dropped from a certain initial height and on which only gravity is acting? And you are asking about the instantaneous velocity when they have reached some final height yf, right?

You are right that yo drops out, but the key point is that the time that the will have taken for each object will be different. For the one dropped from a larger height, the time taken to reach the final position yf will be larger. That's why the velocity will be larger for that object.
 
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nrqed said:
The question is a bit confused. If I understand correctly, you are considering two objects dropped from a certain initial height and on which only gravity is acting? And you are asking about the instantaneous velocity when they have reached some final height yf, right?

You are right that yo drops out, but the key point is that the time that the will have taken for each object will be different. For the one dropped from a larger height, the time taken to reach the final position yf will be larger. That's why the velocity will be larger for that object.

Thank you! That’s what I reasoned out before I confused my self.

This assignment just has me stressed out, I didn’t learn anything in this class, it’s so hard to follow what he’s doing and now I have to do dimensionless analysis all of a sudden. Oof
 
starstruck_ said:
Thank you! That’s what I reasoned out before I confused my self.

This assignment just has me stressed out, I didn’t learn anything in this class, it’s so hard to follow what he’s doing and now I have to do dimensionless analysis all of a sudden. Oof
You are welcome. Although I am a bit puzzled since you mention SR in your title but the formula you use is non relativistic.
 
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nrqed said:
You are welcome. Although I am a bit puzzled since you mention SR in your title but the formula you use is non relativistic.

We haven’t started actual SR yet, we’ve been doing dimensionless analysis to go from the Galilean theory to the Newtonian theory over the past 3 weeks (and I absolutely do not understand anything, or at least I didn’t until I worked my way through part of this assignment).

EDIT: it’s just practice with kinematics formulas we already know
 
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starstruck_ said:
We haven’t started actual SR yet, we’ve been doing dimensionless analysis to go from the Galilean theory to the Newtonian theory over the past 3 weeks (and I absolutely do not understand anything, or at least I didn’t until I worked my way through part of this assignment).

EDIT: it’s just practice with kinematics formulas we already know
Ah ok, that makes sense now :-) Best luck. And don't hesitate to ask questions here!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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