Question

Two particles move in a uniform gravitational field with an acceleration $g=10m/s^2$. At the initial moment the particles were located at one point and moving with velocities $v_1=4.0m/s$ and $v_2=4.0m/s$ horizontally in opposite directions. If the distance(in meters) between the particles at the moment when their velocity vectors become mutually perpendicular is $x$. Find $x$. (Round off your answer to the nearest integer)

Answer 1

Let the velocities of the particles (say ${\overrightarrow{v_1}}^{\prime }$ and ${\overrightarrow{v_2}}^{\prime }$) becomes mutually perpendicular after time $t$. Then their velocities become
${\overrightarrow{v_1}}^{\prime }=\overrightarrow{v_1}+\overrightarrow{g}t$; ${\overrightarrow{v_2}}^{\prime }=\overrightarrow{v_2}+\overrightarrow{g}t$ (1)

As ${\overrightarrow{v_1}}^{\prime }\perp {\overrightarrow{v_2}}^{\prime }$ so,

${\overrightarrow{v_1}}^{\prime }\cdot {\overrightarrow{v_2}}^{\prime }=0$

or, $(\overrightarrow{v_1}+\overrightarrow{g}t)\cdot (\overrightarrow{v_2}+\overrightarrow{g}t)=0$

or $-v_1{v}_2+g^2{t}^2=0$

Hence, $t=\frac{\sqrt{v_1{v}_2}}{g}$
Now form the equation $\overrightarrow{r_{12}}={\overrightarrow{r_0}}_{\left(12\right)}+{\overrightarrow{v_0}}_{\left(12\right)}t+\frac{1}{2}{\overrightarrow{w}}_{12}{t}^2$

$|\overrightarrow{r_{12}}|=|{\overrightarrow{v_0}}_{\left(12\right)}|t$, (because here $\overrightarrow{w_{12}}=0$ and ${\overrightarrow{r_0}}_{\left(12\right)}=0$)

Hence the sought distance

$|\overrightarrow{r_{12}}|=\frac{v_1+v_2}{g}\sqrt{v_1{v}_2}$ (as $|{\overrightarrow{v_0}}_{\left(12\right)}|=v_1+v_2$)

substituting values $v_1=4m/s;v_2=4m/s;g=10m/s^2$

$x=2.8m\simeq 3$