physics

A)Eddie the Eagle, British Olympic ski jumper, is attempting his most mediocre jump yet. After leaving the end of the ski ramp, he lands downhill at a point that is displaced 53.0 m horizontally from the edge of the ramp. His velocity just before landing is 26.0 m/s and points in a direction 37.0$^\circ$ below the horizontal. Neglect any effects due to air resistance or lift.
What was the magnitude of Eddie's initial velocity as he left the ramp?
B)Determine Eddie's initial direction of motion as he left the ramp, measured relative to the horizontal.
C)Calculate the height of the ramp's edge relative to where Eddie landed.

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asked by Sam
  1. V = 26m/s[-37o]

    A. Xo = 26*cos(-37) = 20.76 m/s.

    B. Xo * T = 53 m.
    20.76 * T = 53
    T = 2.55 s. = Fall time.

    Y = 26*sin(-37) = -15.65 m/s.
    Y = Yo + g*T = -15.65
    Yo + 9.8*2.55 = -15.65
    Yo = -15.65 - 24.99 = -40.64 m/s.=Vert.
    component of initial velocity.

    tan A = Yo/Xo = -40.64/20.76 = -1.95761
    A = -63o = 63o below the hor.= Direction

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    posted by Henry

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