Rank the vertical component of velocities at the landing points

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Homework Help Overview

The discussion revolves around ranking the vertical components of velocities for projectiles at the moment they return to the ground. The problem is conceptual in nature, focusing on understanding the relationship between height and vertical velocity without numerical values.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore different rankings of vertical components, with some suggesting that the heights of the projectiles are the primary factor influencing their velocities upon landing. There is confusion regarding how time and height interact in determining these velocities.

Discussion Status

Several participants express uncertainty about their reasoning and the implications of height on vertical velocity. Some suggest that the time taken to reach the ground affects the velocity, while others assert that height is the only relevant factor. There is an ongoing exploration of these ideas without a clear consensus.

Contextual Notes

Participants are navigating a conceptual homework assignment that does not allow for numerical solutions, which may contribute to the confusion regarding the relationships between time, height, and velocity.

isukatphysics69
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Homework Statement


Rank the vertical component of the velocity of each projectile at they moment they return to the ground, greatest first.

Homework Equations


n/a

The Attempt at a Solution


This is a conceptual homework, so there are no numbers involved. But looking at the picture i was thinking
A>B>D>C but it was wrong. I have thought about and tried numerous different approaches but i am completely stuck.
The initial vertical component rankings are B>A=C>D but i don't see how this matters much because at the vertex they will all be 0. The times taken to reach the ground are D>C>A>B So i feel like since D was the quickest it will not have much time to accumulate -9.8m/s each second. I don't know how to approach this problem!
 

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I agree w/ you, although rather than A, B, C, D, I would say, A&B (which are equal) then C&D (which are equal). Who says this is wrong?
 
phinds said:
I agree w/ you, although rather than A, B, C, D, I would say, A&B (which are equal) then C&D (which are equal). Who says this is wrong?
Hi, this is an online homework assignment and it is marking my entry wrong. So why do you say that A&B and C&D are equal for their vertical components? i must be missing something fundamental here.
 
phinds said:
I agree w/ you, although rather than A, B, C, D, I would say, A&B (which are equal) then C&D (which are equal). Who says this is wrong?
Your answer was correct. How did you reason that?
 
I am confused because they all reach that point at different times. I thought that since they all have different times that they reach that point, they will all accumulate a different amount of -9.8m/s and all have different vertical components. How does the height matter?
 
isukatphysics69 said:
I am confused because they all reach that point at different times. I thought that since they all have different times that they reach that point, they will all accumulate a different amount of -9.8m/s and all have different vertical components. How does the height matter?
The question asks about the vertical component so the height is the ONLY thing that matters.
 
phinds said:
The question asks about the vertical component so the height is the ONLY thing that matters.
ok great, thank you!
 
phinds said:
The question asks about the vertical component so the height is the ONLY thing that matters.
To amplify on this there are [at least] two ways you can think about this.

1. The time taken to fall from the peak of a trajectory depends on the height of that peak. The velocity acquired depends on that time alone. So height of the peak is all that matters.

2. [This one is trickier]. The Pythagorean theorem says that the square of the horizontal velocity plus the square of the vertical velocity equals the square of the resulting total velocity. Energy is proportional to the square of velocity. So one can speak loosely of the kinetic energy corresponding to the horizontal velocity and the kinetic energy corresponding to the vertical velocity separately. The sum of the two is total kinetic energy. Horizontal velocity for a freely falling projectile is unchanging. So we can ignore the portion of kinetic energy associated with horizontal motion. The kinetic energy associated with the vertical motion at impact is identical to the gravitational potential energy at the apex of the trajectory. So the vertical velocity at impact is purely a function of the height of the peak of the trajectory.
 

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