February 1, 2023

In “Seek-Lock-Strike!” Again, we obtained the missile’s trajectory. Namely,

$x = \frac{1}{2} \left(\frac{(b-y)^{r+1}}{b^r \cdot f \cdot (r+1)}-\frac{b^r \cdot f \cdot (b-y)^{1-r}}{1-r}\right) -k\quad\quad\quad(*)$

where

$f = \frac{a}{b}+\sqrt{1+(\frac{a}{b})^2},$

$k = \frac{b}{2}\left(\frac{1}{f \cdot (r+1)}-\frac{f}{1-r}\right).$

Since the fighter jet maintains its altitude ( $y= b$ ), the missile must strike it at $(x_*, b)$ . Setting $y=b$ in $(*)$ gives $x_* = -k.$

Hence, we can plot $(*):$

Fig. 1 $a = 100\;m, b=3000\;m, v_a=1500\;ms^{-1}, v_m=2000\;ms^{-1}$

We can also illustrate “Seek-Lock-Strike” in an animation:

Fig. 2 $a = 100\;m, b=3000\;m, v_a=1500\;ms^{-1}, v_m=2000\;ms^{-1}$

Fig. 3 $a = -2500\;m, b=3000\;m, v_a=1500\;ms^{-1}, v_m=2000\;ms^{-1}$

Day: February 1, 2023