An aircraft is flying at a height of $3400 \mathrm{m}$ above the ground. The angle subtended at a ground observation point by the aircraft positions $10.0 \mathrm{s}$ apart is $30°$, what is the speed of the aircraft?

1. Projectile motion:

(i) Time of flight $T= \frac{2u\mathrm{s}\mathrm{i}\mathrm{n}\theta }{g}$

(ii) Horizontal range $R= \frac{u^2\mathrm{s}\mathrm{i}\mathrm{n}2\theta }{g}$

(iii) Maximum height $H= \frac{u^2\mathrm{s}\mathrm{i}{\mathrm{n}}^2\theta }{2g}$

(iv) Trajectory equation (equation of path): $y= x\mathrm{t}\mathrm{a}\mathrm{n}\theta \text{-}\frac{g{x}^2}{2u^2\mathrm{c}\mathrm{o}{\mathrm{s}}^2\theta }= x\mathrm{t}\mathrm{a}\mathrm{n}\theta \left(1\text{-}\frac{x}{R}\right)$

(v) Velocity after time $t$, $v= u\cos \theta \hat{i}+\left(u\sin \theta -gt\right)\hat{j}$

2. Relative motion:

(i) vAB (velocity of A with respect to B) $= v_A-v_B$

(ii) aAB (acceleration of A with respect to B) $= a_A-a_B$

3. Crossing river:

(i) A man in a river always moves in the direction of resultant of velocity of man and velocity of river.

(ii) Velocity with which man moves along the river ${\left({\mathrm{V}}_{\mathrm{m}}\right)}_x= {\mathrm{V}}_{\mathrm{r}}-{\mathrm{V}}_{\mathrm{m},\mathrm{R}}\mathrm{sin\theta }$

(iii) Velocity with which man moves perpendicular to the river ${\left({\mathrm{V}}_{\mathrm{m}}\right)}_{\mathrm{y}}= {\mathrm{V}}_{\mathrm{m},\mathrm{R}}\mathrm{cos\theta }$

(iv) Time take to cross the river, $\mathrm{t}= \frac{\mathrm{W}}{{\mathrm{V}}_{\mathrm{m},\mathrm{R}}\mathrm{cos\theta }}$

(v) To get minimum time $\frac{\mathrm{d}\mathrm{t}}{\mathrm{d}\mathrm{\theta }}= 0\Rightarrow \mathrm{\theta }= 0$

(vi) Drift in this case, $d= \frac{{\mathrm{V}}_{\mathrm{r}}\mathrm{W}}{{\mathrm{V}}_{\mathrm{m},\mathrm{R}}}$

(vii) For shortest path, drift must be zero in this case, and it happens when $\Rightarrow \mathrm{sin\theta }= \frac{{\mathrm{V}}_{\mathrm{r}}}{{\mathrm{V}}_{\mathrm{m},\mathrm{R}}}$