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Zero Launch Angle: Vox and Angle

  1. Dec 5, 2007 #1

    A basketball is thrown horizontally with an initial speed of 4.20 m/s. A straight line drawn from the release point to the landing point makes an angle of 30.0° with the horizontal. What was the release height?

    Vox=4.20 m/s=Vx Ax=0
    Voy=0 m/s Ay=-9.81 m/s^2
    Theta=180*-30.0* = 150*

    2. Relevant equations

    tan [Theta] = (Vy/Vx)

    3. The attempt at a solution

    Tan[theta] x Vx = Vy
    Tan 150* x 4.20 m/s = Vy = -2.42 m/s

    Vy^2 = Voy^2 + 2a[delta]y; Voy=0
    Vy^2 / 2a = [delta]y

    (-2.42 m/s)^2 / -2(9.81m/s^2) = [delta]y = -0.298m

    Which gives a height = 0.298m. So I must have done something wrong????

    back of the book says answer should be 1.20m
    Last edited: Dec 5, 2007
  2. jcsd
  3. Dec 5, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    That should be in terms of distances, not speeds: tan(theta) = Y0/x.

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