Projectile Problem

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  • #1
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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.
 

Answers and Replies

  • #2
Simon Bridge
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If the bullet is fired horizontally, why does it fall to the ground?
 
  • #3
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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 >>
 
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  • #4
Simon Bridge
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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.
 
  • #5
berkeman
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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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