Calculating Horizontal Projectile Displacement and Speed

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Homework Help Overview

The problem involves calculating the horizontal displacement and speed of a projectile fired horizontally with an initial speed of 50 m/s, neglecting air resistance. The specific questions focus on the magnitude of displacement after 3 seconds and the speed of the projectile at that time.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply kinematic equations to find the displacement and speed but questions the initial vertical velocity component. Some participants clarify the meaning of "horizontally" in relation to initial velocities.

Discussion Status

Participants are exploring the implications of the initial conditions, particularly the initial vertical velocity being zero. There is a recognition of the need for clarity on the definitions of horizontal motion, and some guidance has been offered regarding the interpretation of angles and components.

Contextual Notes

The discussion includes a focus on the definitions of horizontal motion and the implications for initial velocity components. There is an acknowledgment of potential confusion regarding the application of trigonometric functions in similar problems.

hakojackie
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The problem states; A projectile is fired horizontally with an initial speed of 50m/s. neglect air resistance. What is the magnitude of the displacement of the projectile 3 seconds after it is fired? What is the speed of the projectil 3 seconds after it is fired? I used this equation to get my X component: X=Vox(t)+(1/2)(-9.8m/s^2)(t^2) and I got 105.9m. I also did this for the Y component I used 50m/s for initial velocity and I think that is what my mistake is, because when I use the pathagorean theorem I don't get the correct answer. Can anyone help?
 
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It's fired horizontally, so what is the initial x velocity? What is the original y velocity?
 
I got the answer finally the initial velocity for Y is 0m/s. I knew that is where I was going wrong but I still don't really understand why it is 0.
 
Because that's what "horizontally" means. If it had any velocity at all in the y direction, either up or down, then it would not be moving horizontal.

It's similar to your other question where it said "60 degrees above the horizontal". "Horizontal" means "0 degrees above the horizontal". You can even do trig on it if you want:

Your other problem:
initial velocity = 30 m/s
x-component = cos(60)*30 = 15 m/s
y-component = sin(60)*30 = 25.98 m/s

This problem
initial velocity = 50 m/s
x-component = cos(0)*50 = 50 m/s, since cos(0)=1
y-component = sin(0)*50 = 0 m/s, since sin(1)=0.

Did you get the right answer? This one was trickier than your last problem.
 

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