Physics

A ball is thrown from a cliff upward with a speed of 27 m/s and at an angle of 71 degrees to the horizontal. How high does it go above the cliff to the nearest tenth of a meter?
answer is 33.3.

I am using 27sin71.

  1. πŸ‘ 0
  2. πŸ‘Ž 0
  3. πŸ‘ 26
asked by Jay
  1. That is the initial velocity in the y-direction.

    Use the following formula:

    Vf^2=Vi^2+2gd

    Where

    d=?
    Vi=(27m/s*Sin71ΒΊ)+2(-9.8m/s^2)d
    Vf=0
    g=-9.8m/s^2

    Solve for d,

    0=(27m/s*Sin 71ΒΊ)^2+2(-9.8m/s^2)d

    0=651.7m/s-19.6m/s^2*d

    (-651.7m/s/-19.6m/s^2)=d

    d=33.25m=33.3m

    1. πŸ‘ 0
    2. πŸ‘Ž 0
    posted by Devron
  2. That is the initial velocity in the y-direction.

    Use the following formula:

    Vf^2=Vi^2+2gd

    Where

    d=?
    Vi=(27m/s*Sin71ΒΊ)
    Vf=0
    g=-9.8m/s^2

    Solve for d,

    0=(27m/s*Sin 71ΒΊ)^2+2(-9.8m/s^2)d

    0=651.7m/s-19.6m/s^2*d

    (-651.7m/s/-19.6m/s^2)=d

    d=33.25m=33.3m

    1. πŸ‘ 0
    2. πŸ‘Ž 0
    posted by Devron

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