3d Projectile Interception

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In the problem, there is a projectile(1) travelling at a constant velocity parallel to the x axis at y= 10. Assuming that at Time 0, projectile 1 is initially at (0,10), If a second projectile(2) were to be fired from (0,0), at an angle [tex]\varphi[/tex] with the y axis, and at an angle [tex]\alpha[/tex] with the z axis, how can angles [tex]\varphi[/tex] and [tex]\alpha[/tex] and the interception time be calculated if projectile 2 was fired at a given time(T), provided:
(A) g = 10m/s^2
(B) [tex]\angle \alpha \le[/tex]45[tex]^\circ[/tex]
(C) Projectile 2 is fired at 50m/s, with negligible air resistance
(D) Projectile 2 is fired to make the fastest possible interception time

[shyamjayakannan]

The time taken to collide with the first projectile is the same as that taken by the second to fall down. So, this time is [tex]t=2\times\frac{v_2\sin\alpha}{g}[/tex]. The horizontal distance covered by the second projectile in this time is [tex]c=v_2\cos\alpha\times t=\frac{2{v_2}^2\sin\alpha\cos\alpha}{g}=\frac{{v_2}^2\sin{2\alpha}}{g}[/tex].

The distance travelled by the first projectile before collision is [tex]a=v_1(T+t)=v_1\left(T+\frac{2v_2\sin\alpha}{g}\right)[/tex].

Now, we know that [tex]c\cos\phi=10 \Rightarrow \frac{{v_2}^2\sin{2\alpha}}{g}\cos\phi=10 \Rightarrow \sin{2\alpha}\cos\phi=\frac{1}{25}\ldots(1)[/tex]
Also, [tex]a=c\sin\phi \Rightarrow v_1\left(T+\frac{2v_2\sin\alpha}{g}\right)=\frac{{v_2}^2\sin{2\alpha}}{g}\sin\phi \Rightarrow T=50\sin{2\alpha}\sin\phi-10\sin\alpha\ldots(2)[/tex]

From [tex](1)[/tex], we get [tex]\cos\phi=\frac{1}{25\sin{2\alpha}} \Rightarrow \sin\phi=\sqrt{1-\frac{1}{25^2\sin^22\alpha}}[/tex] and substituting this in [tex](2)[/tex], we get

[tex]T=50\sin2\alpha\sqrt{1-\frac{1}{25^2\sin^22\alpha}}-10\sin\alpha \Rightarrow (T+10\sin\alpha)^2=50^2\sin^22\alpha\left(1-\frac{1}{25^2\sin^22\alpha}\right)[/tex]

[tex]\Rightarrow T^2+20T\sin\alpha+100\sin^2\alpha=50^2\sin^22\alpha-4\Rightarrow T^2+4+20T\sin\alpha+100\sin^2\alpha=50^2\left(2\sin\alpha\cos\alpha\right)^2[/tex]

[tex]\Rightarrow T^2+4+20T\sin\alpha+100\sin^2\alpha=50^2\times2^2\sin^2\alpha\left(1-\sin^2\alpha\right)[/tex]

[tex]\boxed{\Rightarrow T^2+4+20T\sin\alpha-9900\sin^2\alpha+10000\sin^4\alpha=0}[/tex]

Solve the above equation to get [tex]\sin\alpha[/tex]. We need the smallest positive root because that will minimize the time. Using the root, you can find [tex]\alpha[/tex] and then [tex]\phi[/tex].