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Projectile Motion - Two fired at different times

  1. Mar 2, 2016 #1
    1. The problem statement, all variables and given/known data

    Two projectiles are fired, one at t = 0 with velocity v01, making an angle θ1 with x-axis, and the other at t = T > 0 with velocity v02 making an angle θ2 with x-axis. Show that if they are to collide in midair the travel interval T between the two firings must be:

    [tex] \frac{2v_{01}v_{02}sin(θ_1-θ_2)}{g(v_{01}cos(θ_1) + v_{02}cos(θ_2))}[/tex]


    2. Relevant equations

    [tex]sin(θ_1-θ_2) = sin(θ_1)cos(θ_2)-cos(θ_1)sin(θ_2)[/tex]
    [tex]\frac{d^2x}{dt^2} = 0, \frac{dx}{dt} = v_0x, x = v_{0x}t[/tex]
    [tex] \frac{d^2z}{dt^2} = -g, \frac{dz}{dt} = v_{0z} - gt, z = v_{0x}t - \frac{gt^2}{2}[/tex]
    [tex]\vec r = x \vec i + z \vec j = v_{0x}t \vec i + (v_{0x}t - \frac{gt^2}{2}) \vec j [/tex]


    3. The attempt at a solution

    I used the above equations and the proper indices to create the following equations:

    [tex] v_{01}cos(θ_1)t = v_{02}cos(θ_2)T[/tex]
    [tex] v_{01}sin(θ_1)t - \frac{gt^2}{2} = v_{02}sin(θ_2)T - \frac{gT^2}{2}[/tex]

    I then arranged these so that each equation equaled 0, and then equated them to each other. I got the following equation:

    [tex] v_{01}cos(θ_1)t - v_{02}cos(θ_2)T = v_{01}sin(θ_1)t - \frac{gt^2}{2} - v_{02}sin(θ_2)T + \frac{gT^2}{2}[/tex]

    Simplifying it gave me:

    [tex] v_{01}cos(θ_1)t - v_{02}cos(θ_2)T - v_{01}sin(θ_1)t + v_{02}sin(θ_2)T = \frac{g}{2} (T^2 - t^2)[/tex]

    [tex]\frac{2v_{01}v_{02}(t\frac{(cos(θ_1)-sin(θ_2))}{v_{02}})+T(\frac{(sin(θ_2)-cos(θ_2))}{v_{01}})}{g} = T^2 - t^2 [/tex]

    I don't know where to go from here to get to the solution, or if my process is correct. I tried taking the norm of r and then equating |r1| = |r2|, but that just created a mess with a lot of terms that wouldn't cancel.

    Thanks!







     
    Last edited: Mar 2, 2016
  2. jcsd
  3. Mar 2, 2016 #2

    cnh1995

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    You are asked to verify the expression for t-T. Rearrange this equation to separate t-T on one side. Then substitute t=x/vo1cosθ1 and T=x/vo2cosθ2 on the other side
    and you'll get the desired expression.
     
    Last edited: Mar 2, 2016
  4. Mar 2, 2016 #3

    haruspex

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    You seem to be constantly confused in this question about how to express the time information in the equations.
    At time t > T, how long has the second particle been travelling?
     
  5. Mar 2, 2016 #4
    t-T

    Would I then find t and T from x and plug that into z?
     
  6. Mar 2, 2016 #5

    cnh1995

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    First particle travels for time t, second one travels for time T. You are asked to find the expression for t-T i.e. the interval between the two firings.
     
  7. Mar 2, 2016 #6

    haruspex

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    No, you have it backwards. T is the interval between firings.
     
  8. Mar 2, 2016 #7

    haruspex

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    Your first attempt was fine except that on the right hand side of your first two equations you had T where you should have had t-T. Try that again.
     
  9. Mar 2, 2016 #8
    Should my left side then be [itex] t^2-T^2[/itex]?
     
  10. Mar 2, 2016 #9

    cnh1995

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    Assuming t and T to be the times of travel, I got the desired expression. Maybe I didn't read the qusation properly. But even if I assume so, it won't affect the final expression since there is no time term in it and I'm finding the difference t-T.
     
  11. Mar 2, 2016 #10

    cnh1995

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    If you are assuming T to be the time difference between firings, you can replace t with t1 and T with t2 in your attempt. t1-t2 will be your T. So, your left hand side should be t12-t22.
     
  12. Mar 2, 2016 #11

    haruspex

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    Not sure what you are referring to. In your very first two equations of your attempt, you have t (only) on the left and T (only) on the right.
    The left hand sides are fine; t is the time the first particle has been travelling at time t. Just replace all the Ts on the right with t-T, since that is the time the second particle has been travelling at time t.
     
  13. Mar 2, 2016 #12

    haruspex

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    Sure, but it was going to be hellish confusing with different advice based on different definitions of the same symbols. Assuming cnstntcnfsn read the question correctly, you were outvoted.
     
  14. Mar 2, 2016 #13
    I replaced them, but I am a little confused (I don't know why I can't get this as it seems simple). From the first equation I got [tex] \frac{v_{01}cos(\theta_1)t}{v_{02}cos(\theta_2)} = t - T[/tex]
    I then put this into the 2nd equation and got [tex]v_{01}sin(\theta_1)t = \frac{v_{02}sin(\theta_2)v_{01}cos(\theta_1)t}{v_{02}cos(\theta_2)}+g\frac{T^2}{2}[/tex]
    I can then get [tex] t(v_{01}sin(\theta_1)v_{02}cos(\theta_2)-v_{02}sin(\theta_2)v_{01}cos(\theta_1) )=t v_{01}v_{02}sin(\theta_1-\theta_2)= g\frac{T^2}{2}[/tex]

    Does that look on the right track?
     
  15. Mar 2, 2016 #14

    haruspex

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    You have two unknowns, t and T, but you are only interested in T. So you should use one equation (better not the quadratic one) to express t as a function of T, then use that to get rid of t in the other equation.
     
  16. Mar 2, 2016 #15
    Got it! Thank you so much for your help
     
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