Answer
We found the position vector $\vec r$, as you see in the graph below.
Work Step by Step
First of all, we need to recall the given, which are:
$\bullet$ $v_i=16.0\;\rm m/s$ whereas $v_i$ is the initial speed of the burrito launched from the ground level.
$\bullet$ $\theta_0$ is the initial releasing angle of the burrito launched from the ground level which we change many times as mentioned in the problem.
$\bullet$ $\left(\theta_0\right)_a=40.0^\circ$ whereas $\left(\theta_0\right)_a$ is the first releasing angle.
Now to find the vector $\vec r$ for the releasing first angle of $40^\circ$.
We know that $\vec r$ has two components.
The first one is the horizontal displacement component which is given by
$$x=x_0+v_{ix}t+\frac{1}{2}a_xt^2$$
in which we know that;
the initial (the releasing point) point is zero $x_0=0\;\rm m$, $v_{ix}=v_0\cos\theta_0$, and $a_x=0\;\rm m/s^2$ since we assume that there are no forces acting on the object horizontally.
Thus,
$$x= v_{0}\cos\theta_0t \tag 1$$
The second one is the vertical displacement component which is given by
$$y=y_0+v_{iy}t+\frac{1}{2}a_yt^2$$
in which we know that;
the initial (the releasing point) point is also zero since the burrito is released from level ground $y_0=0\;\rm m$, $v_{iy}=v_0\sin\theta_0$, and $a_y=-g\;\rm m/s^2$ since the object is moving under the free fall acceleration due to Earth's gravitational pull. The negative sign is due to the force direction which is downward toward the Earth's center
Thus,
$$y= v_{0}\sin\theta_0t-\frac{1}{2}gt^2\tag 2$$
Recall that the position vector is given by
$$r=\sqrt{\left(x-x_0\right)^2+\left(y-y_0\right)^2} $$
and we know that, at $t=0$, $x_0=y_0=0$.
Hence,
$$r=\sqrt{\left(x \right)^2+\left(y \right)^2} $$
Plugging from (1) and (2);
$$r=\sqrt{\left( v_{0}\cos\theta_0t \right)^2+\left(v_{0}\sin\theta_0t-\frac{1}{2}gt^2\right)^2} $$
Now we just have to plug the given, the first given and the third given, from above.
$$r=\sqrt{\left( 16\cos40^\circ t \right)^2+\left(16\sin40^\circ t-\frac{1}{2}\cdot 9.8t^2\right)^2} $$
Now we just have to plug this formula into any software calculator to find the vector $\vec r$.
