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Determining Projection Angle With Little Information

  1. Oct 26, 2014 #1
    1. The problem statement, all variables and given/known data

    The speed of a projectile when it reaches its maximum height is one-half its speed when it is at half its maximum height. What is the initial projection angle of the projectile?

    2. The attempt at a solution

    First I tried to find the speed for the max height.
    vf2 - vi2 = -2gΔd
    vf2 = vi2- 2gΔd
    0= vi2- 2gΔd <---
    Can someone explain why this must be set to zero?
    ∴ vi=√(2gh)

    After that I did the same for half the height

    vf2 - vi2 = -2g(Δd/2)
    vf2 = vi2 - g(Δd/2)
    0= vi2 - g(Δd/2)
    ∴ vi2=√(gh)


    After reaching this point I'm not really sure how the given information can help give me an angle. I know somehow I'm supposed to break it down into horizontal components but I just can't see it. Any help would be appreciated.
     
  2. jcsd
  3. Oct 26, 2014 #2

    haruspex

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    That would follow if vf=0, right? Will it be zero at maximum height? What will be zero at maximum height?
     
  4. Oct 26, 2014 #3

    phinds

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    Forgetting the math for the moment, think about this. You are assuming, as one always correctly does in such basic problems, that air resistance is irrelevant and that the trajectory is ballistic. You also know that it's taking off and landing from the same height. What does all this tell you about the horizontal component of the "speed" at the max height vs at the launch? What does THAT tell you about the "speed" at the launch?

    And, I assume that you understand the relationship between speed and velocity, yes?
     
  5. Oct 26, 2014 #4
    oh, let me guess, the final velocity will be zero right? The reason being there's that moment where it's absolutely still in the air.
     
  6. Oct 26, 2014 #5

    haruspex

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    If fired straight up, yes. Is this projectile fired straight up?
     
  7. Oct 26, 2014 #6
    Wouldn't that mean the horizontal component would remain constant, more specifically, the velocity of the horizontal motion wouldn't change.
     
  8. Oct 26, 2014 #7
    No it's an object in projectile motion. But what I'm getting at is that if the object is going on a projectile curve, for example a cannonball shot from a cannon. Wouldn't there be a moment in that parabolic motion where the velocity of the vertical component go from positive to negative? Doesn't that imply that the max height has been reached when that value of velocity is zero?
     
  9. Oct 26, 2014 #8

    phinds

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    Yes and yes. Why would you expect otherwise. What would make it change?
     
  10. Oct 26, 2014 #9

    haruspex

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    Yes, but only the vertical component. So what does vi represent in your equation 0= vi2- 2gΔd?
    The question refers to speeds. So what equations can you write down now?
     
  11. Oct 26, 2014 #10
    the equation 1/2*(final velocity - initial velocity)
    since 0= vi2- 2gΔd is the initial velocity

    then isolate for the final velocity?
     
  12. Oct 26, 2014 #11

    haruspex

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    That's not an equation.
    No, vi is not "the initial velocity" in that equation in this context. Let's step back a moment. What is the 0 in that equation representing? Be exact.
     
  13. Oct 26, 2014 #12
    well, I know vf2 - vi2 = -2gΔd represents the motion from height zero to the max height.
    0= vi2- 2gΔd, so I believe the 0 here would be representing the fact that the vi and 2gΔd are equal right?
     
  14. Oct 26, 2014 #13

    haruspex

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    No, the equation represents that fact. The 0 is the value of a variable. What variable? Be as precise as you can.
     
  15. Oct 26, 2014 #14
    alright, first off thanks for not just spoon feeding me answers. I haven't thought through physics like this in a while.
    So being as literal as I can. 0= vi2- 2gΔd, looking at this equation, the first thing I can think of what that zero represents is the value of vf. So i assume it follows that vi2- 2gΔd point at which vf =0 .If that isn't right, thanks for trying. As of right now it's late where I live and I have classes tomorrow. I'll be back tomorrow to try and work through this. Thanks a lot though.
     
  16. Oct 26, 2014 #15

    haruspex

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    I'm glad you're taking my approach in a good spirit!
    But I don't think we're going to get there at this rate. I've been trying to get you to echo back to me my statement that vf in this equation represents the vertical component of the velocity at maximum height. On that understanding, what does vi represent in that equation?
     
  17. Oct 27, 2014 #16
    That would represent the horizontal component of the velocity at max height
     
  18. Oct 27, 2014 #17

    PeroK

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    I think a diagram of the motion would help a lot. This is a hard problem in my opinion and I would try to think out what's happening (without the equations of motion to begin with). So, I would draw a diagram showing:

    The projectile as it leaves the ground; the projectile at its maximum height; the projectile when it has reached half its maximum height. I'd show the vertical and horizontal components of velocity at each of these points.

    The next step is then to use the equations of motion to work out the velocity and/or speed at each point.

    Also, as you have three points of interest here, it might be better to use ##v_0, \ v_1\ \ v_2## rather than ##v_i, \ v_f##. In fact, keeping track of notation is a bit of a challenge in itself with this problem.
     
    Last edited: Oct 27, 2014
  19. Oct 27, 2014 #18

    haruspex

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    No, no! In the vi, vf notation, i stands for initial and f for final. As PeroK says, if there are multiple initial and final speeds to be considered (different directions, different phases of the action) it is a very good idea to use subscripts or a different letter to discriminate those. In this case, how about v0, v1, v2 for the vertical speeds (v2 being 0) and u (which does not change) for the horizontal speed?
    In a given SUVAT equation, all the distances and speeds are in the same direction, so you apply the equations separately in (usually) the vertical and horizontal directions.
    So please write down some equations relating to the vertical movement here.
     
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