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Math Help - derivation of range formula when initial height is differnt from final

  1. #1
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    derivation of range formula when initial height is differnt from final

    What is the derivation for this formula?
    R=(1+\sqrt{1-\frac{2gy}{v_o^2sin^2\theta_o}})\frac{v_o^2}{2g}si  n2\theta_o
    R=horizontal displacement
    y=final height-initial height
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by superdude View Post
    What is the derivation for this formula?
    R=(1+\sqrt{1-\frac{2gy}{v_o^2sin^2\theta_o}})\frac{v_o^2}{2g}si  n2\theta_o
    R=horizontal displacement
    y=final height-initial height
    Take a look at me


    Give it a healthy reading...till maximum range ...take care of difference of signs in your question and in Wikipedia's proof...If you still find any trouble feel free to ask.

    Adarsh
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