Answer
$3.7\;\rm m$
Work Step by Step
First of all, we need to find the time of the trip from the end of the conveyor to the top of the pipe.
$$y_f=y_i+v_{iy}t+\frac{1}{2}a_yt^2$$
We know that the initial height is 3 m, (which means that the final height is zero meters since we chose the top of the pipe to be our origin). The sand moves at the free-fall acceleration.
$$0=3+6\sin(-15^\circ)t-\frac{1}{2}gt^2$$
$$4.9t^2+6\sin15^\circ t-3=0$$
Thus, $t=-0.957\;\rm s$ (rejected) and $t=0.6399\;\rm s$
Now we can find $d$ since we knew that the horizontal velocity component is constant and we also knew the time it takes.
$$d=x_f-x_i=v_{ix}t=v_i\cos(-15)t=6\cos(15)\cdot 0.6399$$
$$\boxed{d=\color{red}{\bf3.7}\;\rm m}$$
