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Projectile Motion!

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose a stone is thrown from a 45.0 m tall building at an angle 26.0° below the horizontal. If it strikes the ground 30.3 m away, find the following values. (a) time of flight

    (b) initial speed
    _________ m/s

    (c) speed and angle of the velocity vector with respect to the horizontal at impact
    _________m/s ______°

    2. Relevant equations
    vix = vicostheta
    viy = visintheta

    dx = vxt
    d = vit + 1/2at2

    3. The attempt at a solution
    I tried to find vix and viy in terms of vi and then used dx = vxt to get t as 30.0m/.899vi but I don't know if (1) I'm doing it right and (2) where to go from here.
  2. jcsd
  3. Nov 17, 2008 #2


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    Homework Helper

    You should get 2 equations:

    30.3 = V*Cosθ * t

    45 = V*Sinθ * t + 1/2 * g* t2

    V * t = 30.3/Cosθ

    ==> 45 = Sinθ * (30.3/Cosθ ) + 1/2*g*t2

    1/2*g*t2 + 30.3*Tanθ - 45 = 0

    Solve the quadratic for t and then figure V from either equation.

    For final velocity Vfy = V*Sinθ + g*t Your Vfx is still V*Cosθ so you should be done.
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