Answer
We found the position vector $\vec r$, as shown in the graph below, from this formula;
$$\boxed{r=\sqrt{\left( 16\cos80^\circ t \right)^2+\left(16\sin80^\circ t-\frac{1}{2}\cdot 9.8t^2\right)^2}} $$
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.
$\bullet$ $\left(\theta_0\right)_b=80.0^\circ$ whereas $\left(\theta_0\right)_a$ is the first releasing angle.
Now to find the vector $\vec r$ for the releasing first angle of $80^\circ$.
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.
$$\boxed{r=\sqrt{\left( 16\cos80^\circ t \right)^2+\left(16\sin80^\circ t-\frac{1}{2}\cdot 9.8t^2\right)^2}} $$
Now we just have to plug this formula into any software calculator to find the vector $\vec r$.
