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Thread: projectile

  1. #1
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    projectile

    a particle is projected from a point on the horizontal ground with a speed u inclined to a angle a to the horizontal ground. the particle is at a height of h above the horizontal ground at times t1 and t2, show that t1*t2=2h/g
    this is what ive tried h=usinat-4.9t^2 i tired putting that into the -b forumula but got no where
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  2. #2
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    Re: projectile

    For a projectile launched and landing at ground level, the total time of motion is $t_f=\dfrac{2v_0\sin(a)}{g} = \dfrac{2v_{0y}}{g}$

    Since the projectile is at the same height, $h$, at times $t_1$ and $t_2$, then $t_1-t_0=t_f-t_2 \implies t_1+t_2=t_f+t_0$.

    Assuming launch occurs at $t_0=0$ ...

    $\color{red}{t_1+t_2=\dfrac{2v_{0y}}{g}}$

    $(t_1+t_2)^2 = t_1^2+2t_1t_2+t_2^2= \dfrac{4v_{0y}^2}{g^2} \implies$

    $\color{red}{t_1^2+t_2^2=\dfrac{4v_{0y}^2}{g^2}-2t_1t_2}$

    Now add the two equations for height ...

    $h=v_{0y} \cdot t_1 - \dfrac{1}{2}gt_1^2$
    $h=v_{0y} \cdot t_2 - \dfrac{1}{2}gt_2^2$
    -------------------------------------------------------
    $2h = v_{0y}(t_1+t_2)-\dfrac{1}{2}g(t_1^2+t_2^2)$

    Substitute the expressions for $\color{red}{t_1+t_2}$ and $\color{red}{t_1^2+t_2^2}$ and solve for $t_1t_2$.
    Thanks from markosheehan
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    Re: projectile

    when i am subbing it in do you know what t1t2 stands for . i cant find an expression for it
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    Re: projectile

    i am trying to sub it in this is where i get to

    2h=usina(2usina/g) -4.9(4u^2 sina^2-2t1t2) dont know what t1t2 is. do you know if im doing it right
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  5. #5
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    Re: projectile

    Quote Originally Posted by markosheehan View Post
    when i am subbing it in do you know what t1t2 stands for . i cant find an expression for it
    $t_1 \cdot t_2$ is just the product of the two times the projectile is at height $h$ ...

    I worked out two expressions for you ...

    $\color{red}{t_1+t_2=\dfrac{2v_{0y}}{g}}$ and $\color{red}{t_1^2+t_2^2 = \dfrac{4v_{0y}^2}{g^2} - 2}\color{blue}{t_1t_2}$

    I also summed the two equations for same height at the two times ...

    $2h = v_{0y}\color{red}{(t_1+t_2)} - \dfrac{1}{2}g\color{red}{(t_1^2+t_2^2)}$

    substitute ...

    $2h = v_{0y} \cdot \color{red}{\dfrac{2v_{0y}}{g}} - \dfrac{1}{2}g \cdot \color{red}{\left(\dfrac{4v_{0y}^2}{g^2} - 2\color{blue}{t_1t_2}\right)}$

    $2h = \dfrac{2v_{0y}^2}{g} - \dfrac{2v_{0y}^2}{g} + g(t_1 \cdot t_2)$

    $2h = \cancel{\dfrac{2v_{0y}^2}{g}} - \cancel{\dfrac{2v_{0y}^2}{g}} + g(t_1 \cdot t_2)$

    $\dfrac{2h}{g} = t_1 \cdot t_2$

    ... which is what the problem asked you to show.
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    Re: projectile

    just in case you want to see it i came up with a much easier way to work this out let t be the 2 times t1 and t2 the 2 equations are h=ut-.5gt^2 and h=ut-.5gt^2 add this together. 2h=2ut-gt^2 rearrange gt^2-2ut+2h=0 divide by g t^2-2ut/g+2h/g t1 and t2 are the roots so t1*t2=-2h/g
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  7. #7
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    Re: projectile

    ?? You can't "let t be the two times t1 and t2". They are two different times, not the same time.
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  8. #8
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    Re: projectile

    T can be 2 different times as it's a quadratic. I just should not of named it that way. It's still a valid way of working out the question
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