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Why is V_x of a projectile constant?

  1. Jul 2, 2015 #1
    1. The problem statement, all variables and given/known data

    Why is the horizontal velocity for a projectile constant?

    I understand it physically, since the vertical and horizontal velocities aren't related and don't affect each other, g will only affect V_y and neglect V_x, so it's constant.

    But maths-wise, isn't V_x = V*Cos(theta)? How can it be constant if the angle is changing?

    2. Relevant equations

    V_x = V*Cos(theta)

    3. The attempt at a solution
     
  2. jcsd
  3. Jul 2, 2015 #2

    rude man

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    V is also changing, such that Vcosθ is constant.
    Good question!
     
  4. Jul 2, 2015 #3

    PeroK

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    I
    ##\theta## refers to the initial angle at launch, not the changing angle at which the projectile moves thereafter.
     
    Last edited: Jul 2, 2015
  5. Jul 2, 2015 #4
    But the law for the horizontal velocity is V_x = V*Cos(theta) where theta is the current angle the velocity vector makes with the x+ axis, right?
     
  6. Jul 2, 2015 #5

    PeroK

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    No, that's wrong. ##v## is the magnitude of the initial velocity and ##\theta## is the initial angle. That's why ##v_x## is constant.
     
  7. Jul 2, 2015 #6

    SammyS

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    That's only true if θ is defined that way. Clearly OP considers θ to be changing along the trajectory.

    In this case we would likely θ0 as the initial value of θ,

    As usual, rude man has answered well. - - short and complete.
     
  8. Jul 2, 2015 #7

    PeroK

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    Apologies, my mistake. I see what the OP was asking.
     
  9. Jul 2, 2015 #8

    rude man

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    Thanks Sammy. Owe you one!
     
  10. Jul 2, 2015 #9

    rude man

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    You are correct, and vcosθ is constant unless you try to include air friction. Which you don't want to do, trust me! :smile:
     
  11. Jul 2, 2015 #10
    Okay, thanks.

    Another question, it's not worth making a thread for.


    If the velocity of a projectile is given by V = 25i -4.9j (y is directed upwards), has the projectile reached its highest point yet? The book says yes.


    Since it's moving downwards, that means yes. But what if I initially threw it down, and it bounced back up higher than its starting point?
     
  12. Jul 2, 2015 #11

    SammyS

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    That's no longer simple projectile motion.
     
  13. Jul 2, 2015 #12

    rude man

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    Then its V would no longer be i 25 - j 4.9.
     
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