Answer
$297.6\;\rm m$
Work Step by Step
To find $d$, in the figure below, which is the distance down the slope at which the arrow hits the ground, we need to find point $B=(x_1,y_1)$.
We chose point $A=(x_0,y_0)=(0,0)$ to be the origin and hence,
$$b=\sqrt{x_1^2+y_1^2}\tag 1$$
We know that the $v_x$ is constant while $v_y$ changes since $a_y=-g$.
Thus,
$$y_1=1.75+v_{0y}t-\frac{1}{2}gt^2$$
$$y_1=1.75+50\sin20^\circ t-4.9t^2\tag 2$$
$$x_1=v_{x0}t$$
$$x_1=50\cos20^\circ t\tag 3$$
From the geometry of the figure below, we can see that
$$\tan(-15^\circ)=\dfrac{y_1}{x_1}$$
Plugging from (2) and (3);
$$\tan(-15^\circ)=\dfrac{1.75+50\sin20^\circ t-4.9t^2}{50\cos20^\circ t}$$
Solving for $t$;
$t=-0.05838$ s, or $t=6.118$ s.
We have to dismiss the negative root.
Now we need to plug (2) and (3) into (1) and substitute by $t=6.118$ s.
$$d=\sqrt{\left[50\cos20^\circ \cdot 6.118\right]^2+\left[1.75+(50\sin20^\circ \cdot 6.118) -(4.9\cdot 6.118^2)\right]^2}$$
$$\boxed{d=\color{red}{\bf297.6}\;\rm m}$$
