Muzzle Vel. Calc: Proj. Problem (vx)

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Homework Help Overview

The discussion revolves around calculating the muzzle velocity of a projectile based on its launch and landing coordinates. The problem is situated within the context of projectile motion, specifically examining horizontal and vertical components of motion.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use the Pythagorean theorem to find the muzzle velocity but expresses confusion regarding the parabolic trajectory of the projectile. Some participants question the assumptions about the bullet's motion, particularly why it falls to the ground if fired horizontally.

Discussion Status

Participants are exploring different aspects of the problem, including the forces acting on the bullet in both the x and y directions. There is an emphasis on guiding the original poster towards understanding the principles of projectile motion without providing direct solutions. The conversation reflects a mix of theoretical discussion and attempts to clarify the problem setup.

Contextual Notes

There is a note regarding the rules of the forum, emphasizing that participants should not do the homework for the original poster but rather guide them in their understanding. This highlights the importance of the learning process in the discussion.

vanitymdl
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Your gun's muzzle is at coordinate (0,1.12) meters and
Your projectile hit the ground at coordinate (1.03,0) meters

Compute muzzle velocity (vx):This is all I'm given to figure out the muzzles velocity, so I know I'm given a height and how much it moved horizontally. But I'm getting confused to how I need to find the vx. I was simply going to do the Pythagorean theorem,
But I don't think that would make much sense because the graph should look more like a parabola.
 
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If the bullet is fired horizontally, why does it fall to the ground?
 
bullet problem

The force on the bullet in the x direction is Fx = m(d2x/dt2) = 0.
The force on the bullet in the y direction is Fy = m(d2y/dt2) – mgy = 0.
Integrating the first equation twice w.r.t. time we get
x = << Too much detailed help deleted by Mentors >>
 
Last edited by a moderator:
The force on the bullet in the y direction is Fy = m(d2y/dt2) – mgy = 0.
You just wrote that the force on the bullet in the y direction is zero. You know that right?

Note: it is against the rules to do homework problems for people.
The idea is to guide them to discovering how to do it for themselves.
 
f.wright said:
The force on the bullet in the x direction is Fx = m(d2x/dt2) = 0.
The force on the bullet in the y direction is Fy = m(d2y/dt2) – mgy = 0.
Integrating the first equation twice w.r.t. time we get
x = << Too much detailed help deleted by Mentors >>

Please remember that the student must do the bulk of the work. You are welcome to provide hints, find errors, ask probing questions, etc. But it is against the PF rules to do the student's homework for them.
 

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