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Projectile Motion man tosses object

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

    A man tosses object with initial velocity of 20 m/s into a room, and is 2.0 m above the ground. The room is 10.0 m above the ground.

    In a provided diagram, the distance from man to room horizontally is 31.8 m.

    I am required to solve for the angle of the object as it leaves the mans hand.

    2. Relevant equations

    Big 5


    3. The attempt at a solution

    vix= 20cosθ
    viy= 20sinθ

    To solve for θ, I must use the initial velocity given, broken up into components.

    Since I am given the vertical displacement (8.0 m) and horizontal displacement (31.8 m) , I can use 2 formulas and substitute one into the other, by isolating t as it is common to both.

    dx= vixt
    31.8 = 20cosθt
    t= 31.8 / 20cosθ

    dy= viyt + 0.5ayt2
    8.0 = 20sinθt - 4.9t2
    8.0 = 20sinθ(31.8 / 20cosθ) - 4.9(31.8 / 20cosθ)2
    8.0 = (31.8sinθ / cosθ) - 4.9(2.5281 / cos2θ)
    8.0 = (31.8sinθ / cosθ) - (12.39 / cos2θ)

    and now i'm stuck.... I tried moving to one side and changing to common denominator...

    8.0 = (31.8sinθ / cosθ) - (12.39 / cos2θ)

    0 = (31.8sinθ / cosθ) - (12.39 / cos2θ) - 8.0

    0 = (31.8sinθcosθ - 12.39 - 8.0cos2θ) / cos2θ

    0 = 31.8sinθcosθ - 12.39 - 8.0cos2θ

    0 = 15.9sin2θ - 12.39 - 8.0(1 - sin2θ)

    0 = 15.9sin2θ - 12.39 - 8.0 + 8sin2θ

    0 = 8sin2θ + 15.9sin2θ - 20.39

    but no luck...

    i thought it looked like a quadratic at first, but I don't think it works with sin2θ

    any suggestions? thank you :)
     
  2. jcsd
  3. Feb 26, 2014 #2

    Simon Bridge

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    ... so you have interpreted the question as asking about the direction of the initial velocity vector, given the ball passes through points A and B (going from A to B) and you know the speed at A.

    ... trig identities not helping you - you may be able to do it geometrically seeing ##2\sin^2\theta = 1-\cos2\theta## ... or is there another kinematic equation you haven't considered yet - that may be called into play?
    i.e. have you used all the information at your disposal yet?


    Notes:
    do not put the numbers in until the end or you'll get turned around.
    keep track of your reasoning all the way through.
     
  4. Feb 26, 2014 #3

    Simon Bridge

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    Hmmm ... perhaps:
    ... go back to where it looks like you almost have a quadratic in ##\cos\theta##,
    ... multiply through by ##\sec^2\theta## ... from there, Pythagoras should get you a quadratic in ##\tan\theta##.
     
  5. Feb 26, 2014 #4
    Mhm. Didnt even notice that it could be changed to secant. Thankyou
     
  6. Feb 27, 2014 #5

    Simon Bridge

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    No worries.
    The tan vs sec identities crop up a lot - especially in exams.
     
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