Answer
$$\mathbf{r}(t)=600 \sqrt{3} \ t \mathbf{i}+(600 \ t -16 \ t^2) \mathbf{j} \mathbf{k}$$
Work Step by Step
Given$$ \mathbf{r}^{\prime \prime}(t)=-32 \mathbf{j}, \quad \mathbf{r}^{\prime}(0)=600 \sqrt{3} \mathbf{i}+600 \mathbf{j}, \quad \mathbf{r}(0)=\mathbf{0}$$
As we have $$\mathbf{r^{\prime}}(t)=\int \mathbf{r}^{\prime \prime}(t) dt$$
and
$$\mathbf{r^{}}(t)=\int \mathbf{r}^{\prime }(t) dt$$
So, integrating on a component-by-component basis produces
\begin{array}{l}
{\mathbf{r^{\prime}}(t)=\int \mathbf{r}^{ \prime\prime}(t)(t)dt=\int\left(-32 \mathbf{j}\right) d t= -32 t\mathbf{j}+\mathbf{C_1}} \\
\text{since} \ \mathbf{r^{\prime}}(0)=600 \sqrt{3} \mathbf{i}+600 \mathbf{j}, \quad \text{we get}
\\ {\mathbf{r^{\prime}}(0)= 0 \mathbf{j}+\mathbf{C_1}=600 \sqrt{3} \mathbf{i}+600 \mathbf{j}}
\\ { \Rightarrow \mathbf{C_1}=600 \sqrt{3} \mathbf{i}+600 \mathbf{j}} \\
{\text {so we get }}\\
{\mathbf{r^\prime}(t)=-32t\mathbf{j}+600 \sqrt{3} \mathbf{i}+600 \mathbf{j}=600 \sqrt{3} \mathbf{i}+(600-32t) \mathbf{j}}\\
\\
{\text{Also we have}}\\
{\mathbf{r}(t)=\int \mathbf{r}^{\prime}(t)(t)dt=\int\left(600 \sqrt{3} \mathbf{i}+(600-32t) \mathbf{j}\right) d t\\
= 600 \sqrt{3} \ t \mathbf{i}+(600 \ t -16 \ t^2) \mathbf{j}+\mathbf{C}} \\
\text{since} \ \mathbf{r}(0)=0, \quad \text{so we get}
\\ {\mathbf{r}(0)= 0 \mathbf{i}+ 0\mathbf{j}+\mathbf{C}=0}
\\ { \Rightarrow \mathbf{C}=0} \\
{\text {so we get }}\\
{\mathbf{r}(t)=600 \sqrt{3} \ t \mathbf{i}+(600 \ t -16 \ t^2) \mathbf{j}} \mathbf{k}
\end{array}
