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Projectile Motion

  1. Nov 22, 2008 #1
    1. A boy can throw a stone at 14 m/s. At what angle must he throw it in order to hit a window which is 10m away horizontally and 5m above the ground?



    2. I would assume we have to use trig ratios and the kinematics equations.

    v = u + a.t
    v^2 = u^2 + 2.a.s
    s = u.t + 1/2.a.t^2





    3. Diagram:

    img.png
    (see attachment or http://img377.imageshack.us/my.php?image=imgwn9.png )


    I know that time is a constant for both the horizontal and vertical components, so I presume that it would not affect the equations.

    My info:
    -Vertical motion
    (t=0)
    s = 5m
    u = 14 m/s sin.(theta)
    a = +9.8 m/s^2

    -Horizontal motion
    (t=0)
    s = 10m
    u = 14 m/s cos.(theta)
    a= 0

    So how do I find theta? Am I missing any equations? Is my diagram correct? Is u = 14 m/s sin.(theta) and u = 14 m/s cos.(theta) correct? Thanks :)

    I'm thinking of using some equation in theta... but what those equations are is beyond me =[

     
    Last edited: Nov 22, 2008
  2. jcsd
  3. Nov 22, 2008 #2
    you should write how horizontal and vertical distances depend on t, then express t from one equation (horizontal makes it easier), and put it in other.
     
  4. Nov 22, 2008 #3

    Mentallic

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    Homework Helper

    There is more manipulation of trigonometric identities involved in this question than actually using formulas.
    You have the horizontal velocity vector. Now you want to find how long it will take the projectile to hit the window. This wont be a direct answer, but in terms of [tex]\theta[/tex].
    Use [tex]t=\frac{s_x}{u_x}[/tex]
    where,
    sx=horizontal displacement
    ux=horizontal velocity vector

    Then from here you can substitute the time into:
    [tex]s_y=u_yt+\frac{1}{2}a_yt^2[/tex]
    where,
    sy=vertical displacement
    uy=vertical velocity vector
    ay=acceleration due to gravity (make sure this is negative, since it is acting opposite to the velocity vector and the displacement) i.e. -9,8ms-2

    You will now find that you have an equation all in terms of [tex]\theta[/tex]. Now comes the manipulation. You will be surprised to find that there are two possible angles to throw the projectile at :smile:
     
  5. Nov 22, 2008 #4

    Mentallic

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    If you're curious to know, I made some calculations and found for this scenario, but instead being under the influence of stronger gravity (being on another planet); the gravity can only go as high as [tex]g=\frac{49(1-\sqrt{5})}{5} \approx -12.11ms^{-2}[/tex] until the rock would never be able to reach the window. (not taking into account that higher gravity will slow the initial velocity of the throw).
    At this gravity, the precise angle to throw the rock at would be [tex]tan^{-1}(\frac{\sqrt{5}+1}{2}) \approx 58^o17'[/tex]
     
  6. Nov 23, 2008 #5
    t = s/u
    s = 10m
    u = 14cos.[tex]\theta[/tex] m/s [horizontal velocity vector]

    so, t = 10 / 14cos.[tex]\theta[/tex]

    then, s = u.t + 1/2.a.t^2

    5 = (14sin[tex]\theta[/tex] * 10/14cos[tex]\theta[/tex]) + 1/2.(-9.8).(10/14cos[tex]\theta[/tex])(10/14cos[tex]\theta[/tex])

    Please excuse the absence of units- including them would make it so much more confusing.

    I will deal with each part of that equation separately:


    ut:
    (14sin[tex]\theta[/tex] * 10/14cos[tex]\theta[/tex])
    = 140sin[tex]\theta[/tex] / 14cos[tex]\theta[/tex]
    Right? I don't know how to continue frome here...

    1/2 * a:
    1/2.(-9.8)
    = -.49

    t^2
    (10/14cos[tex]\theta[/tex])(10/14cos[tex]\theta[/tex])
    = 100 / 14cos^2[tex]\theta[/tex]
    Don't know how to continue from here either...

    So, I have:

    5 = (140sin[tex]\theta[/tex] / 14cos[tex]\theta[/tex]) + (-4.9)(100 / 14cos^2[tex]\theta[/tex])

    I guess my problem now is working with the trig identities now...
     
  7. Nov 23, 2008 #6

    Mentallic

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    ok first we will simplify the equation (just the coefficients at first. i.e. the numbers).

    [tex]5=14sin\theta(\frac{10}{14cos\theta})+\frac{1}{2}(-9.8)(\frac{10}{14cos\theta})^2[/tex]

    simplified: [tex]5=\frac{10sin\theta}{cos\theta}-4.9(\frac{25}{49cos^2\theta})[/tex]

    Now the trig identities you should be aware of are as follows. If you want to know the proofs for them, just ask :smile:

    1) [tex]tan\theta=\frac{sin\theta}{cos\theta}[/tex]
    thus [tex]tan^2\theta=\frac{sin^2\theta}{cos^2\theta}[/tex]

    2) [tex]sin^2\theta+cos^2\theta=1[/tex]

    3) [tex]sec^2\theta=1+tan^2\theta[/tex] N.B. [tex]sec\theta=\frac{1}{cos\theta}[/tex]

    See if you can apply these formulas in a variety of ways to get an equation all in terms of the same trig ratio. tangent, cosine or sine. Good luck!
     
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