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Math Help - Another projectile questions involves distance

  1. #1
    Junior Member silvercats's Avatar
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    Another projectile questions involves distance

    Object is projected from a speed of V and an angle of Θ
    Horizontal displacement is 15m ,vertical displacement is 5m
    Find V and Tan Θ

    (right )
    x=V cos Θ*t
    t=x/V cos Θ

    (up)
    y=v Sin Θ*t-g/2*t
    from first equation y=V sinΘ(x/V cosΘ)-g/2*(x/v cosΘ)
    5=V sin Θ(15/V cos Θ)-g/2 *(15/V cosΘ) <<<Stuck!!!!
    help wanted

    Books says 12.5 , 4/3
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by silvercats View Post
    Object is projected from a speed of V and an angle of Θ
    Horizontal displacement is 15m ,vertical displacement is 5m
    Find V and Tan Θ

    (right )
    x=V cos Θ*t
    t=x/V cos Θ

    (up)
    y=v Sin Θ*t-g/2*t
    instead of using 't' you should have used 't/2' in the above eqn. do you see why?
    from first equation y=V sinΘ(x/V cosΘ)-g/2*(x/v cosΘ)
    5=V sin Θ(15/V cos Θ)-g/2 *(15/V cosΘ) <<<Stuck!!!!
    help wanted

    Books says 12.5 , 4/3
    did this help?
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  3. #3
    Junior Member silvercats's Avatar
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    Quote Originally Posted by abhishekkgp View Post
    instead of using 't' you should have used 't/2' in the above eqn. do you see why?
    why?????
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  4. #4
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by silvercats View Post
    why?????
    't' is the total time of flight, that is the time taken by the projectile to hit the ground. it takes half the time of flight to achieve maximum height(here max height is 5m). now do you understand what i was trying to say?
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  5. #5
    Junior Member silvercats's Avatar
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    ah.. yeah.can you guide me through the rest?
    thanks
    Last edited by silvercats; May 26th 2011 at 10:02 PM.
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