1. With an awning of length A at right angles to the canvas and the canvas vertical, no rain drops will hit the canvas if vc / vr = A/H. This follows directly by looking at the figure.

2. The rain velocity vr is the vertical component of the velocity of the rain, while the wind moves with velocity vw. By the Pythagorean theorem therefore, the rain moves at a speed of sqrt(vr2 + vw2) with respect to the ground. These are illustrated in the following figures where the canvas of height H moves from left to right with an awning of length A. Segment T hits the bottom horizontal axis at the leftmost point at which it could hit without causing rain to hit the bottom of the canvas. That point is the base of the vertical segment B.

First, the canvas will be tilted forward in the same way as would be required for zero wind velocity. The triangles HBD and the one formed by velocities vc and vr are similar; therefore,

D = H vc /sqrt(vc2 + vr2 )
B = H vr / sqrt(vc2 + vr2 )

Secondly, it is clear from the cartoon that the optimum position of the swiveled awning is such that the drop trajectory T at maximum wind velocity (just missing the bottom point p of the canvas) is a tangent to the circle with the radius A and with the center at the upper end of the canvas. The triangles TEF and BTA are similar because the angle between A and T is 90 degrees as is the angle between E and F. Also, the angle between F and T is 90 minus the angle between B and T and therefore it is equal to the angle between A and B. This implies that the angle between B and T is equal to the angle between T and E. Hence all three angles are the same in the two triangles so TEF and BTA are similar, when the wind pushes the rain as far left as possible while still missing the canvas. Further the triangle TEF and the one formed by velocities vw and vr is similar. Intuitively, the awning takes care of the wind velocity vw whereas the incline of the painting takes care of the movement velocity vc. This intuition is tempered only by the fact that the incline of the painting determines the length of B. The smaller B, the less the danger the canvas will be hit.

 
So, we have
A/B = F/T
F/T = vw /sqrt(vw2 + vr2 )
A = B F/T = H vr vw /sqrt{(vc2 + vr2 ) (vw2 + vr2 )}.
(vc2 + vr2 ) (vw2 + vr2 ) = (H vr vw / A)2
vc2 (vw2 + vr2 ) = (H vr vw / A)– vr2 (vw2 + vr2 )
vc2 = ((H vr vw / A)2 /(vw2 + vr2 ) )  – vr2
 
From that it follows
v(A/H) vr sqrt{(vc2 + vr2 )/[ vr2 - (A/H)2(vc2 + vr2 )]}
 
Similarly, the canvas velocity is given as
vc = vr sqrt{( vwH /A) 2 /(vw2 + vr2 ) – 1}

So, suppose the raindrops are falling 5 meters per second, the height is 3 meters, and the awning A is 0.8 meters off the canvas. How fast should you move if the headwind velocity may range from 0 up to 1.5 meters per second? What angle should the canvas be to the ground? What angle should the awning be to the canvas?

You have to move at least 2 meters per second (about twice a normal walking pace). If so, the canvas will be tilted about 22 degrees from the vertical and the awning will be about 95 degrees from the canvas (but still slanted downwards to the right). Moving faster will force a greater tilt which will allow you to tolerate a greater windspeed. Move fast, stay dry.
 
The solutions are due to Ivan Rezanka.