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Homework Help: Launch angle of projectile

  1. Jan 26, 2007 #1
    1. The problem statement, all variables and given/known data
    At .5 of its maximum height, the speed of a projectile is .75 of its initial speed. What was its launch angle?

    2. Relevant equations
    (1)V1y = V0y - gt
    (2)y1-y0 = V0yt - .5gt^2

    3. The attempt at a solution
    The part that really bothers me is the second part of the first statement: "the speed of a projectile ..." What is the speed of a projectile? What do i denote that as? V?...
    So far I've gotten the following:
    initial velocity y-component:
    knowing: V1y = 0 (at max height)
    V1y = V0y - gt
    Voy = gt
    substitute V0y into equation (2) gives max height:

    Any assistance would be most appreciated.

    Last edited: Jan 26, 2007
  2. jcsd
  3. Jan 26, 2007 #2


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    Hi un0rth0d0x,
    welcome to PF. Hope we can help you here with you physics questions.

    Since the projectile is subject to just a conservative force this problem can be solved using energy considerations.

    Have you done the relevant theory concerning energy conservation yet?
  4. Jan 26, 2007 #3
    Unfortunately I have not... This problem is sectioned in basic kinematics... So i'm totally oblivious to energy conservation.

    Last edited: Jan 26, 2007
  5. Jan 26, 2007 #4


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    The "speed" of the projectile [tex]v[/tex] has two components [tex]v_x,\ v_y[/tex] of which the y-component changes and the x-components stays the same throughout the trajectory.
  6. Jan 26, 2007 #5
    andrevdh: just out of interest, how would you sort this out using energy considerations alone? I have had a quick bash, and can't see an obvious way to do it...
    anyway, I would approach this problem with kinematic equations of motion.
    In situations like this, remember that if you're given letters or proportions then you effectively "know" the value, don't get scared just because there isn't an actual number. It seems that's exactly what you've done, so good on you.
    Second, you should realise that any projectile motion problems involving no air resistance are simply parabolic geometry problems in disguise
    third, always but always draw a diagram! it helps to visualise the problem.
    Now, if you have the x components and the y components, what shape can you make with them?
  7. Jan 26, 2007 #6


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    As we know the speed of a projectile changes during its motion. The problem therefore says that the speed [tex]v[/tex] (the x- and y- components combined) at half maximum height will be 3/4 of the launching speed [tex]v_0[/tex].
  8. Jan 26, 2007 #7


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    Remember that the x - velocity component can be written as

    [tex]v_x = v_o \cos(\theta _o)[/tex]

    while we need to solve for

    [tex]\theta _o[/tex]
  9. Jan 26, 2007 #8


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    energy at bottom:

    [tex]0.5mv_o ^2[/tex]

    energy in middle:

    [tex]\frac{9}{32}mv_o ^2 + 0.5mgh[/tex]

    energy at top:

    [tex]0.5mv_o ^2 {cos}^2(\theta_o) + mgh[/tex]
  10. Jan 26, 2007 #9


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    The problems says that the speed at half the maximum height is 3/4 of the initial speed. Since the speed of the projectile is the combination of both speed components we can say that

    [tex]v^2 = v_y ^2 + v_x ^2[/tex]

    at half the maximum height this therefore becomes

    [tex](\frac{3v_o}{4})^2 = v_y ^2 + v_o ^2 {cos}^2(\theta_o)[/tex]
  11. Jan 26, 2007 #10


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    Can you solve it now?
  12. Jan 26, 2007 #11
    Thanks for your help!

  13. Jun 18, 2009 #12
    Sorry to bump such an old thread.

    I am trying to figure out this exact question, and I see what steps they have taken in this thread, but I don't understand where to go from there. Any additional assistance would be appreciated, I am just plain stuck on this.

  14. Jun 18, 2009 #13


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    (\frac{3v_o}{4})^2 = v_y ^2 + v_o ^2 {cos}^2(\theta_o)

    Now we have one equation with two unknowns, so try to express vy in terms of v0. Hint: use vf^2 - vi^2 = 2ad. Twice.
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