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cnstntcnfsn
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Homework Statement
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]
Homework 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]
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!
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