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physics

posted by on .

please show me how to do this in steps.

A track star in the long jump goes into the jump at 12 m/s and launches herself at 20.0degrees above the horizontal. What is the magnitude of her horizontal displacement?
(assume no air resistance and that
Ay= -g = -9.81 m/s^2)

thanks :)

  • physics - ,

    Her vertical component of velocity at takeoff is
    Vyo = 12 sin 20 = 4.10 m/s
    and her horizontal component is
    Vx = 12 cos 20 = 11.28 m/s.
    The horizontal component remains constant during the jump.
    Use the vertical velocity component to determine how long she is in the air before hitting the ground. Call that time T

    Vo = g T/2
    T = 2 Vo^2/g

    The length of the jump is Vx*T
    = 2 (Vo^2/g) sin 20 cos 20
    = (Vo^2/g) sin 40

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