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Math Help - veloctity of a ball.

  1. #1
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    veloctity of a ball.

    Can anyone please guide/help me with this question, i have attached it to the post,
    thank you.

    The equation looks so complicated, I just dont know where to even begin?
    Attached Thumbnails Attached Thumbnails veloctity of a ball.-screen-shot-2013-02-26-17.22.13.png  
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: veloctity of a ball.

    Quote Originally Posted by Tweety View Post
    Can anyone please guide/help me with this question, i have attached it to the post,
    thank you.

    The equation looks so complicated, I just dont know where to even begin?
    Hi Tweety!

    Let's begin with v=0.
    I believe you have an expression for v.
    Can you substitute it?
    And then rewrite it to find t?
    Thanks from Tweety and topsquark
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  3. #3
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    Re: veloctity of a ball.

    Thank you, I got  t= tan^{-1}\frac{\alpha V_{n}}{g} ?

    Is that correct? Do i use this value to fine ymax
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  4. #4
    Super Member ILikeSerena's Avatar
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    Re: veloctity of a ball.

    Quote Originally Posted by Tweety View Post
    Thank you, I got  t= tan^{-1}\frac{\alpha V_{n}}{g} ?

    Is that correct? Do i use this value to fine ymax
    You should have:

    t = \frac 1 {\alpha \sqrt g} \tan^{-1} \frac {\alpha v_n} {\sqrt g}

    This is the time that v will be zero, and y will be on its maximum.
    The time t depends on the coefficient \alpha.

    But yes, then you use this value to find y_{max}.
    Thanks from Tweety
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  5. #5
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    Re: veloctity of a ball.

    Quote Originally Posted by ILikeSerena View Post
    You should have:

    t = \frac 1 {\alpha \sqrt g} \tan^{-1} \frac {\alpha v_n} {\sqrt g}

    This is the time that v will be zero, and y will be on its maximum.
    The time t depends on the coefficient \alpha.

    But yes, then you use this value to find y_{max}.
    So I put this value into the second equation?

    Which is very complicated, I dont know how to simplify it, I put t = ot the above , and got this

     y = \frac{1}{\alpha} ln(cos(tan^{-1}\frac{\alpha v_{n}}{\sqrt{g}} - \alpha\sqrt{g} \frac{1}{\alpha \sqrt{g}}) tan^{-1}\frac{\alpha V_{n}}{\sqrt{g}}

    I dont know how to simplify this
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  6. #6
    Super Member ILikeSerena's Avatar
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    Re: veloctity of a ball.

    Quote Originally Posted by Tweety View Post
    So I put this value into the second equation?

    Which is very complicated, I dont know how to simplify it, I put t = ot the above , and got this

     y = \frac{1}{\alpha} ln(cos(tan^{-1}\frac{\alpha v_{n}}{\sqrt{g}} - \alpha\sqrt{g} \frac{1}{\alpha \sqrt{g}}) tan^{-1}\frac{\alpha V_{n}}{\sqrt{g}}

    I dont know how to simplify this
    It looks like you did not properly substitute.
    When you properly substitute there is a lot that disappears because it's zero.
    Btw, you could have simplified the fraction with \alpha \sqrt g in it - it's just 1.
    No need to do that now though, because you first need the proper substution.
    Furthermore, if you can't reasonably simplify, that's fine. It means you are done and that you have the answer.
    Thanks from Tweety
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  7. #7
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    Re: veloctity of a ball.

    Oh, okay

    I this time I tried it i got,

     \frac{1}{cos(tan^{-1}\frac{\alpha V_{n}}{\sqrt{g}}} ) is this correct?

    Actually sorry missed out the ln term,

    so I got  lncos(tan^{-1}\frac{\alpha V_{n}}{\sqrt{g}}} )
    Last edited by Tweety; February 27th 2013 at 02:05 AM.
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  8. #8
    Super Member ILikeSerena's Avatar
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    Re: veloctity of a ball.

    Quote Originally Posted by Tweety View Post
    Oh, okay

    I this time I tried it i got,

     \frac{1}{cos(tan^{-1}\frac{\alpha V_{n}}{\sqrt{g}}} ) is this correct?

    Actually sorry missed out the ln term,

    so I got  lncos(tan^{-1}\frac{\alpha V_{n}}{\sqrt{g}}} )
    Much better!
    But you still need a factor \frac{-1}{\alpha^2} in front and then you have y_{max}.


    Btw, if you want you can simplify more.
    That is, you can use that \cos(\tan^{-1}(\frac y x)) = \frac {x} {\sqrt{x^2+y^2}}
    Last edited by ILikeSerena; February 27th 2013 at 02:45 AM.
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