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Math Help - 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 \theta_1, \theta_2, at velocity 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:

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

    Resolving Vertically gives:

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

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

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

    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 t_2.

    How do i simplify that?

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

    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:

    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; November 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 \sin(\theta_1 - \theta_2)?

    I get:

    t_1 = \dfrac{<br />
2V_0\left(\sin(\theta_1) - <br />
  \dfrac{\cos(\theta_1)\sin(\theta_2)}{\cos(\theta_2  )\right)}<br />
  }<br />
{g\left(<br />
  \dfrac{\cos^2(\theta_1)}{\cos^2(\theta_2)}<br />
 -1\right)}<br />
}

    Which then gives:

    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:

    t_1 = \dfrac{2V_0\left(\sin(\theta_1) - \cos(\theta_1)\tan(\theta_2))}<br />
{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 \theta_1, \theta_2, at velocity 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 t_{1} and t_{2}, only the difference between them.

    If you take the equation

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