Answer
$$\mathbf{r}=8 t \mathbf{i}+8 t \mathbf{j}+\left(100-16 t^{2}\right) \mathbf{k}$$
Work Step by Step
Since
$$\frac{d \mathbf{r}}{d t}=\int(-32 \mathbf{k}) d t=-32 t \mathbf{k}+\mathbf{C}_{1}$$ and $$ \frac{d \mathbf{r}}{d t}(0)=8 \mathbf{i}+8 \mathbf{j} \Rightarrow-32(0) \mathbf{k}+\mathbf{C}_{1}=8 \mathbf{i}+8 \mathbf{j} \Rightarrow \mathbf{C}_{1}=8 \mathbf{i}+8 \mathbf{j} $$
Then
$$ \frac{d \mathbf{r}}{d t}=8 \mathbf{i}+8 \mathbf{j}-32 t \mathbf{k}$$
and
\begin{align*}
\mathbf{r}&=\int(8 \mathbf{i}+8 \mathbf{j}-32 t \mathbf{k}) d t\\
&=8 t \mathbf{i}+8 t \mathbf{j}-16 t^{2} \mathbf{k}+\mathbf{C}_{2}
\end{align*}
$$ \mathbf{r}(0)=100 \mathbf{k} \Rightarrow 8(0) \mathbf{i}+8(0) \mathbf{j}-16(0) \mathbf{j}-16(0)^{2} \mathbf{k}+\mathbf{C}_{2}=100 \mathbf{k}$$
Then
$ \mathbf{C}_{2}=100 \mathbf{k}$$$\mathbf{r}=8 t \mathbf{i}+8 t \mathbf{j}+\left(100-16 t^{2}\right) \mathbf{k}$$
