A particle starts from the origin at $t=0s$ with a velocity of $10.0\hat{j} m/s$ and moves in the $x-y$ plane with a constant acceleration of $(8.0\hat{i}+2.0\hat{j}) m{s}^{-2}$. The $x$-coordinate of the particle is $16m$ then the$\frac{(y-\mathrm{coordinate})}{4}$ is  (in m)

Here, $\overrightarrow{u}=10.0\hat{j} m/s$ at $t=0$.

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=(8.0\hat{i}+2.0\hat{j}) m{s}^{-2}$

So, $d\overrightarrow{v}=(8.0\hat{i}+2.0\hat{j}) dt$

Integrating it within the limits of motion i.e., as time changes from $0$ to $t$, velocity changes from $u$ to $v$, we have

$$\begin{gathered} \overrightarrow{v}-\overrightarrow{u}=(8.0\hat{i}+2.0\hat{j})t \\ \overrightarrow{v}=\overrightarrow{u}+8.0t \hat{i}+2.0t \hat{j} \end{gathered}$$

As, $\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt} \Rightarrow d\overrightarrow{r}=\overrightarrow{v}dt$

So, $d\overrightarrow{r}=(\overrightarrow{u}+8.0t \hat{i}+2.0t \hat{j}) dt$

Integrating it within the conditions of motion i.e., as time changes from $0$ to $t$, displacement is from $0$ to $r$, we have,

$$\begin{gathered} \overrightarrow{r}=\overrightarrow{u}t+\frac{1}{2}\times (8.0\hat{i}+2.0\hat{j})t^2 \\ x\hat{i}+y\hat{j}=10t \hat{j}+4.0t^2 \hat{i}+t^2 \hat{j} \\ =4.0t^2 \hat{i}+(10t+t^2)\hat{j} \end{gathered}$$

Here, we have, $x=4.0t^2 ; y=10t+t^2$

$\therefore t={\left(\frac{x}{4}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

 At $x=16 m ; t=\sqrt{\frac{16}{4}}=2s$

 We have, $y=10\left(2\right)+2^2=24 m$