A particle starts from origin at t=0 with a velocity $5\widehat{i}m{s}^{-1}$ and moves in x-y plane under the action of a force which produces a constant acceleration of $3\widehat{i}+2jm{s}^{-2}.$The y-coordinate of the particle at the instant when its x-coordinate is 84m, is  

The position of the particle is given by $\overrightarrow{\mathrm{r}}=\overrightarrow{{\mathrm{r}}_0}+\overrightarrow{{\mathrm{v}}_0}\mathrm{t}+\frac{1}{2}\overrightarrow{\mathrm{a}}{\mathrm{t}}^2$

Where, $\overrightarrow{r_0}$ is the position vector at t=0 and ${\overrightarrow{v}}_o$ is the velocity at t =0

Here, $\overrightarrow{{\mathrm{r}}_0}=0,{\overrightarrow{\mathrm{v}}}_0=5\widehat{\mathrm{i}}{\mathrm{ms}}^{-1},$

$\overrightarrow{\mathrm{a}}=(3\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}){\mathrm{ms}}^{-2}$

$\therefore \overrightarrow{\mathrm{r}}=5\mathrm{ti}+\frac{1}{2}(3\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}){\mathrm{t}}^2;$

$\overrightarrow{\mathrm{r}}=\left(5\mathrm{t}+1.5{\mathrm{t}}^2\right)\widehat{\mathrm{i}}+1{\mathrm{t}}^2\widehat{\mathrm{j}}........\left(\mathrm{i}\right)$

Compare it with $\overrightarrow{r}=x\widehat{i}+y\widehat{j}\text{, we get }\mathrm{X}=5\mathrm{t}+1.5{\mathrm{t}}^2,\mathrm{y}=1{\mathrm{t}}^2$

$\therefore \mathrm{X}=84\mathrm{m}; 84=5\mathrm{t}+1.5t^2$

On solving, we get t=6 s $\text{ At }t=6\mathrm{s},\mathrm{y}=\left(1\right)(6{)}^2=36$