Calculating Velocity Magnitude for Projectile Motion

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Homework Help Overview

The discussion revolves around calculating the magnitude of the final velocity of a projectile, specifically a ball thrown upward from a height with an initial speed and angle. The subject area includes concepts from kinematics and projectile motion.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of initial and final velocity components, questioning the appropriateness of using certain indices for velocity. There is discussion about the application of conservation of energy principles and the formulas available for the problem.

Discussion Status

The discussion is ongoing, with participants providing guidance on the use of indices for velocity and clarifying the application of conservation of energy. Some participants express confusion about the concepts discussed, indicating a need for further clarification.

Contextual Notes

There is mention of limited formulas available to the original poster, and a note that conservation of energy has not been covered in class, which may affect the understanding of the problem.

Suprin
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Homework Statement


A ball is thrown upward from a platform 5.2meters high with a speed of 15m/s at an angle of 40° from the horizontal. What is the magnitude of its velocity when it hits the ground?

distance = 5.2m
V_0 = 15m/s
angle = 40°

Homework Equations


Don't have the answer to this question. Want to make sure I got the right answer.

The Attempt at a Solution



V_0x = 15cos40° = 11.491

V_0y = 15sin40° = 9.642

|V| = \sqrt{11.491^2 + 9.642^2} = 15m/s ?
 
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You got the magnitude of initial velocity. You need to find the final velocity.
 
How about now?

Final velocity on X is equivalent to the initial velocity on X, so that remains the same.

V_fy^2 = (9.642)^2 + 2(9.81)(5.2)

V_fy^2 = 92.968 + 102.024

\sqrt{V_fy^2} = \sqrt{194.992} = 13.964

|V| = \sqrt{11.491^2 + 13.964^2} = 18.084Out of curiosity, did I get the magnitude of the initial velocity right?
 
Last edited:
You can check the initial velocity yourself, it was given :)

Regarding your solution, it seems correct, but. You should not use ##V_{0y}## for the vertical component of final velocity. ##V_0## is already used to mean initial velocity.

Finally, you don't have to use conservation of energy just for the vertical components. You could just as well have used it for total velocity.
 
voko said:
You can check the initial velocity yourself, it was given :)

Regarding your solution, it seems correct, but. You should not use ##V_{0y}## for the vertical component of final velocity. ##V_0## is already used to mean initial velocity.

Finally, you don't have to use conservation of energy just for the vertical components. You could just as well have used it for total velocity.

I literally write down things the same way the professor gave us on our formula sheets.

And you completely lost me on the whole conservation of energy thing. If anything, that hasn't been mentioned during class.
 
The index 0 usually means "initial". You are looking for final velocity, so 1 or f would be more appropriate there, giving ##V_1## or ##V_f##, respectively.

As for the conservation of energy, you used that to obtain the vertical component of the final velocity. If that sounds strange to you, explain what formula and used, and why.
 
voko said:
The index 0 usually means "initial". You are looking for final velocity, so 1 or f would be more appropriate there, giving ##V_1## or ##V_f##, respectively.

As for the conservation of energy, you used that to obtain the vertical component of the final velocity. If that sounds strange to you, explain what formula and used, and why.

Ah, ignore the sub-zero thing. That was a typo. Not used to writing math formulas on the PC.

We haven't worked with conservation of energy yet, nor air resistance or anything like that. The formulas I used (and why) are the ones I have available, for now, to obtain the information that is asked of me to find. It's one of those cases where I have about 4 formulas (for example), and I'm asked to find Time and only one of them has Time involved in it; so I know that's the correct one.
 
Uhm, OK. That formula does not require conservation of energy for its derivation, it just happens to follow most naturally from that. Anyway, your result is correct.
 

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