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Projectile Trajectory

  1. Aug 30, 2015 #1
    1. The problem statement, all variables and given/known data
    An artillery shell is fired at an angle of 23.9◦ above the horizontal ground with an initial speed of 1530 m/s.

    The acceleration of gravity is 9.8 m/s2 .

    Find the total time of flight of the shell, neglecting air resistance.

    Answer in units of min.

    2. Relevant equations

    Vf^2=Vi^2+2ad
    d=Vit+1/2at^2

    3. The attempt at a solution
    I tried using the first equation above with the givens to find distance, however, I don't know final velocity. Once finding the distance I would have plugged it into the second equation above to find the time. I would have then converted the time in seconds to find the time in minutes.

    However, none of this was possible because I didn't know the final velocity.
     
  2. jcsd
  3. Aug 30, 2015 #2

    Bystander

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  4. Aug 30, 2015 #3
    I understand this, however how can you calculate final velocity with just the angle and initial velocity?
     
  5. Aug 30, 2015 #4

    Bystander

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    No idea how to find components of a vector?
     
  6. Aug 30, 2015 #5

    SteamKing

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    Since you are firing the projectile at an elevated angle above the horizontal, what happens to the projectile once it's fired? You can draw a picture of the trajectory if that will help you to visualize what's going on.

    Remember, since you are dealing with gravity, what goes up must come back down. :wink:
     
  7. Aug 31, 2015 #6
    Thank you! So I understand what the projectile looks like. I have drawn a picture and tried a different attempt to what I was doing before. I found the x and y components of the angle. I then used the y component of this answer as the initial vertical velocity, used 0 as the final vertical velocity, and used-9.8 as the vertical acceleration. I then plugged these values into the Vy=Viy+at to find the time in seconds. I got 63.25s. I then converted this to 1.054 minutes. However, this answer was wrong.

    I still don't understand what I am doing wrong. Any ideas?
     
  8. Aug 31, 2015 #7
    Yes, I used the y component of the vector as explained in my reply to "SteamKing" However, I still got the answer wrong.

    Any ideas on where I went wrong?
     
  9. Aug 31, 2015 #8

    SteamKing

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    Did you remember what I told you about gravity? What goes up must come back down?
     
  10. Aug 31, 2015 #9
    Oh right, okay so my time is half of the time that it takes the shell to fly. So I must double my time?
     
  11. Aug 31, 2015 #10

    SteamKing

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    Yes.
     
  12. Aug 31, 2015 #11
    Thank you!
     
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