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Thread: Two balls thrown and collide in mid air, produce an expression for the time between

  1. #1
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    Two balls thrown and collide in mid air, produce an expression for the time between

    Hello!

    Basically I am stuck... been pulling my hair out trying to simplify this and come out with a simple expression!
    Also not sure what area of the forum to put this in, but seeing as there is nothing too complicated and its all algebra i think this is the right place.

    Heres the problem:
    Two balls are thrown at angles $\displaystyle \theta_1$, $\displaystyle \theta_2$, at velocity $\displaystyle V_0$ one after the other such that they collide at point B in mid air, determine the time between the throws (analytically).

    Heres what i have done so far:
    Resolving horizontally gives:

    $\displaystyle t_1 = \frac{\cos(\theta_2) t_2}{\cos(\theta_1)}$
    and
    $\displaystyle t_2 = \frac{\cos(\theta_1) t_1}{\cos(\theta_2)}$

    Resolving Vertically gives:

    $\displaystyle V_0 \sin(\theta_1) t_1 -\frac{g}{2}t^2_1 = V_0 \sin(\theta_2) (\frac{\cos(\theta_1)t_1}{\cos)\theta_2)}) - \frac{g}{2} (\frac{\cos^2(\theta_1) t^2_1}{\cos^2(\theta_2)})$

    Sub in the expression for $\displaystyle t_2$ to find the eqn in terms of $\displaystyle t_1$ then my plan is to simplify and do the same for $\displaystyle t_2$ until finally i can do:

    $\displaystyle \Delta{t} = t_1 - t_2$ where $\displaystyle \Delta{t}$ is the answer to the problem.

    Okay, so heres where i am stuck:
    Trying to simplify the eqn. for $\displaystyle t_1$, im not sure exactly how to do this:

    $\displaystyle t_1 =\frac{2V_0(\frac{\sin(\theta_1-\theta_2)}{\cos(\theta_2)})}{g(1-\frac{\cos^2(\theta_1)}{\cos^2(\theta_2)})} $

    I have a similar expression for $\displaystyle t_2$.

    How do i simplify that?

    EDIT: I am guessing there is a trig identity for bottom part which can be written:

    $\displaystyle g\frac{cos^2(\theta_2) - cos^2(\theta_1)}{\cos^2(\theta_2)}$

    but i cant remember it!

    EDIT 2: Perhaps there isnt, so i tried simplifying and got this:

    $\displaystyle t_1 = \frac{2V_0\sin(\theta_1-\theta_2)\cos(\theta_2)} {g\cos^2 (\theta_2)-\cos^2(\theta_1)}$

    Is that okay?
    Last edited by darksupernova; Nov 8th 2010 at 02:30 AM.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Where did you get $\displaystyle \sin(\theta_1 - \theta_2)$?

    I get:

    $\displaystyle t_1 = \dfrac{
    2V_0\left(\sin(\theta_1) -
    \dfrac{\cos(\theta_1)\sin(\theta_2)}{\cos(\theta_2 )\right)}
    }
    {g\left(
    \dfrac{\cos^2(\theta_1)}{\cos^2(\theta_2)}
    -1\right)}
    } $

    Which then gives:

    $\displaystyle t_1 = \dfrac{2V_0\cos(\theta_2)\left(\sin(\theta_1)cos(\ theta_2) - \cos(\theta_1)\sin(\theta_2))}{g\left(\cos^2(\thet a_1)- \cos^2(\theta_2) \right)}$

    Or you change the previous one to:

    $\displaystyle t_1 = \dfrac{2V_0\left(\sin(\theta_1) - \cos(\theta_1)\tan(\theta_2))}
    {g\left(\cos^2(\theta_1)sec^2(\theta_2) -1\right)}} $
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  3. #3
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    Two balls are thrown at angles $\displaystyle \theta_1$, $\displaystyle \theta_2$, at velocity $\displaystyle V_0$ one after the other such that they collide at point B in mid air, determine the time between the throws (analytically).


    You are not asked to calculate $\displaystyle t_{1}$ and $\displaystyle t_{2}$, only the difference between them.

    If you take the equation

    $\displaystyle V_{0}\sin\theta_{1}t_{1}-(1/2)gt_{1}^{2}=V_{0}\sin\theta_{2}t_{2}-(1/2)gt_{2}^{2},$

    and get just the $\displaystyle gt^{2}$ terms on one side, you can isolate $\displaystyle t_{2}-t_{1},$ (by factorising) and then substitute for the ratio of $\displaystyle t_{1}/t_{2}$ on the other side.
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