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Tangent Line on Projectile

  1. Oct 8, 2015 #1
    • Warning: Members are reminded to use the provided Formatting Template for homework posts!
    My professor did this question in class and I am a little confused. I wrote it down in my notes but I kind of don't understand it.

    The question is: Find theta 1/4 of the way through the flight of a projectile in time

    He does not give us numbers. Everything has to be solved algebraically.

    My question is the angle at 1/4 of the flight the same as the angle of the projectile when it is fired?

    And also is the slope of the tangent line at any point on the projectile the same as the slope at that point on the projectile?
     
    Last edited: Oct 8, 2015
  2. jcsd
  3. Oct 8, 2015 #2

    PhanthomJay

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    Yes, the direction of the projectile at any point in time defines the slope (tangent) of the curve at that point.
    Edit: didn't pay attention to your first question. No to that one. The direction of the projectile is changing at each instant.
     
    Last edited: Oct 8, 2015
  4. Oct 8, 2015 #3

    ehild

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    Welcome to PF!
    See the picture: It is the track of a projectile. Is the angle of the tangent of the trajectory is the same everywhere? It would be a straight line then!
    Yes, the angle of slope of the tangent line is the same as the angle of slope of the velocity vector of the projectile. The velocity is tangent to the track of motion..
    You need the angle the velocity vector encloses with the horizontal direction. How do the horizontal and vertical components of velocity change with time?
    What is the time of the whole flight in terms of the initial velocity?
    upload_2015-10-9_5-13-12.png
     
  5. Oct 9, 2015 #4
    So the slope of the tangent line at that point is equal to the velocity at that point? He doesn't give us numbers. Everything is suppose to be denoted algebraically.

    The total flight is t. My notes look something like that. I am confused about the tan.
     

    Attached Files:

    Last edited: Oct 9, 2015
  6. Oct 9, 2015 #5

    PhanthomJay

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    The slope at time t/4 gives you the direction of the velocity vector at that point. The angle of that slope is calculated as you show : tan theta = Vy/Vx. So you need to find Vy and Vx at that point in time. But that depends on the initial velocity of the projectile at the time it was fired, both it's magnitude and direction, so I assume that was given also as say V_initial at firing angle alpha? Note the firing angle and the angle you are trying to calculate are not the same, so call the firing angle alpha , not theta.
     
  7. Oct 10, 2015 #6

    ehild

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    Assume that the initial velocity is of magnitude Vo and it makes the angle α with the horizontal.
    You know that the motion of the projectile can be treated as a horizontal and a vertical motion. The horizontal one with constant velocity and the vertical one with constant acceleration. What are the horizontal (x) and vertical (y) components of the initial velocity?
    How long is the projectile in air ? Denote it by T.
    How do the components of the velocity depend on time?
    You are right about the tangent of angle theta. If you have the equations for vx and vy, you can write up the equation for tanθ in terms of time, and substitute what you got for t=T/4.
     
  8. Oct 11, 2015 #7
    Is tan used to find the angle?

    I got to:
    tan= sin/cos - gtf/4vcos
     
  9. Oct 11, 2015 #8

    ehild

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    The formula you wrote has no sense.
    tan, sin, cos are functions, have no meaning without their argument. The tangent of an angle is its sine over cosine: [tex]\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}[/tex]
     
  10. Oct 11, 2015 #9
    Yeah, its confusing because he won't give us any numbers. He just did halfway through the problem and said to continue it ourselves.
     
  11. Oct 11, 2015 #10

    ehild

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    Solve the problem symbolically. Call the initial angle alpha, and denote the angle at t=Tfinal/4 with theta.
     
  12. Oct 11, 2015 #11
    You say he started the problem on the board but didn't finish it? There has to be more information given than what you're telling us. What was the original launch angle? Does the projectile fly across level ground and land at the same height it started from?

    I suggest you start over, using the template because it guides you through the start of the process.

    Without that template we can't help you.
     
  13. Oct 12, 2015 #12
    So since tan=vy/vx

    Vx is vcos
    Vy is vsin-g(tf/4)
     
  14. Oct 12, 2015 #13

    haruspex

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    You need to distinguish between the angle of the trajectory at launch and the angle at t/4.
     
  15. Oct 12, 2015 #14
    Why do I need to do that? I am using tan to find the angle at the 1/4.
     
  16. Oct 12, 2015 #15

    haruspex

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    You wrote
    What angle is referred to there by "sin"?
     
  17. Oct 12, 2015 #16
    The angle of 1/4 of projectile.
     
  18. Oct 12, 2015 #17

    haruspex

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    Then why the -gt/4 term?
    (By the way, the projectile is the object. You mean 1/4 of the trajectory.)
     
  19. Oct 12, 2015 #18
    Yes, 1/4 of the trajectory. -gt/4 is 1/4 of the trajectory. He wants it in terms of time.
     
  20. Oct 12, 2015 #19

    haruspex

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    What do you mean "-gt/4 is 1/4 of the trajectory"? It is what at 1/4 of the trajectory?
     
  21. Oct 12, 2015 #20
    I actually dont know. He wrote it dovn to find Vy you would use vf=vi+at and then i sub everything in to get that. Do you know why he used the formula?
     
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