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I have two projectiles that are launched at different times. The one that is launched second should collide with the first one mid-air. Given are the launch angles αP1&αP2, start speeds v0P1&v0P2 and start positions(x0P1/y0P1)&(x0P2/y0P2) of both projectiles. I'm looking for the launch time of the second projectile so that $$ y0_P1+(v0_P1⋅sin(α_P1))⋅t1−(1/2)⋅g⋅t1^2=y0_P2+(v0_P2⋅sin(α_P2))⋅t2−(1/2)⋅g⋅t2^2 $$ $$x0_P1+(v0_P1⋅cos(α_P1))⋅t1=x0_P2+(v0_P2⋅cos(α_P2))⋅t2$$

As projectile is launched after the first, t2=(t1+tOffset). Now I tried to solve the above equations for tOffset and but didn't really get anywhere. Is this the correcty way to get what I'm looking for?

Thanks in advance

Don_M
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1 Answers1

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I suggest eliminating $t$ from the two sets of equations to get an equation of the form $$y=y_0+(x-x_0)\tan\alpha+{g\over2v_0^2\cos^2\alpha}(x-x_0)^2$$ for each parabola. Subtracting one from the other will give you a quadratic equation in $x$ for the intersections of the two parabolas. Assuming that there are any intersections, it’s a simple matter to compute the travel times of the two projectiles to those $x$-coordinates and then find their difference. You’ll need to discard any solution for which either of the times or the difference between the second and first is negative, of course.

amd
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