Question

Two particles are simultaneously thrown from top of two towers as shown. Their velocities are $2\text{m/s}$ and $14\text{m/s}$. Horizontal and vertical separation between these particles are $22\text{m}$ and $9\text{m}$ respectively. Then the minimum separation between the particles in process of their motion in meters is ($g=10{\text{m/s}}^2$)

Answer 1

By resolving into components $\overrightarrow{u}=2\cos {45}^{\circ }\hat{i}+2\sin {45}^{\circ }\hat{j}$
$\overrightarrow{v}=-14\cos {45}^{\circ }\hat{i}+14\sin {45}^{\circ }\hat{j}$

relative velocity along $x-$ axis is $v_x=8\sqrt{2}\text{m/s}$
$\therefore x=(22-8\sqrt{2}t)$
relative velocity along $y-$ axis is $v_y=6\sqrt{2}\text{m/s}$
$\therefore y=(9-6\sqrt{2}t)$

Separation between the particles
$\therefore r=\sqrt{x^2+y^2}$
$r=\sqrt{x^2+y^2}=\sqrt{(22-8\sqrt{2}t{)}^2+(9-6\sqrt{2}t{)}^2}$
$r=\sqrt{{22}^2+(8\sqrt{2}t{)}^2+9^2+(6\sqrt{2}t{)}^2-460\sqrt{2}t}$
$r^2={22}^2+200t^2+9^2+-460\sqrt{2}t$
If $r$ is minimum , $r^2$ is also minimum
$\frac{d{r}^2}{dt}=0\Rightarrow t=\frac{23}{10\sqrt{2}}\text{s}$

Substituting $t$ in equation,
$r_{\text{min}}=6.00\text{m}$