Answer
The $y$ coordinate at that time is
$$
y=45 \mathrm{m} .
$$
Work Step by Step
Given the initial velocity $\vec{v}_{0}=(8.0 \mathrm{m} / \mathrm{s}) \hat{\mathrm{j}}$ and the acceleration
$$\vec{a}=\left(4.0 \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(2.0 \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}},$$
the position vector of the particle is
$$\vec{r}=\vec{v}_{0} t+\frac{1}{2} \vec{a} t^{2}=(8.0 \hat{\mathrm{j}}) t+\frac{1}{2}(4.0 \hat{\mathrm{i}}+2.0 \hat{\mathrm{j}}) t^{2}=\left(2.0 t^{2}\right) \hat{\mathrm{i}}+\left(8.0 t+1.0 t^{2}\right) \hat{\mathrm{j}}$$
Therefore, the time that corresponds to $x=29 \mathrm{m}$ can be found by solving the equation
$2.0 t^{2}=29,$ which leads to $t=3.8 \mathrm{s}$ . The $y$ coordinate at that time is
$$
y=(8.0 \mathrm{m} / \mathrm{s})(3.8 \mathrm{s})+\left(1.0 \mathrm{m} / \mathrm{s}^{2}\right)(3.8 \mathrm{s})^{2}=45 \mathrm{m} .
$$
