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March 25, 2017

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A ball launched from ground level lands 2.15 s later on a level field 50 m away from the launch point. Find the magnitude of the initial velocity vector and the angle it is above the horizontal. (Ignore any effects due to air resistance.)

  • physics - ,

    u = constant horizontal speed = 50/2.5 = 20 m/s

    v = vi - 9.8 t
    at top of arc, t = 2.5/2 = 1.25 seconds and v = 0
    so at top
    0 = vi - 9.8 (1.25)
    so
    vi = 12.25

    so speed at start = sqrt (20^2 + 12.25^2)
    = sqrt(400+ 150) = sqrt 550 = 23.5 m/s

    tan angle = 12.25/20
    so
    angle = 31.5 degrees

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