Physics

A projectile is fired at v0= 355.0 m/s at an angle of 68.7% with respect to the horizontal. Assume that air friction will shorten the range by 32.1%. How far will the projectile travel in the horizontal direction, R?

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  1. time in air:
    hf=h+355*sin68.7*t - 4.9t^2
    t= 355sin68.7/4.9=67.5sec
    dhoriztal=355*67.5*(1-.321)

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    bobpursley
  2. Vo = 355m/s[68.7o].
    Xo = 355*Cos68.7 = 129.0 m/s.
    Yo = 355*sin68.7 = 330.8 m/s.

    Y = Yo + g*Tr = 0,
    330.8 + (-9.8)Tr = 0,
    Tr = 33.80 s. = Rise time.
    Tf = Tr = 33..80 s. = Fall time.

    R = Xo*(Tr+Tf) = 129*67.6 = 8720.4 m. with no air friction.
    R = 8720.4 - 0.321*8720.4 = 5921 m. with air friction.

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