Relative motion on a conveyor belt

[JJBladester]

Homework Statement

Belt A conveys sand with a velocity of 6.0 ft/s (constant). The velocity of belt B is 8 ft/s (constant). Determine the velocity of the sand relative to B as it lands on belt B. Answer: V_S/B=20.1 ft/s at 85.1 degrees

Homework Equations

V_S/B = V_S-V_B
(Velocity of sand relative to belt B = velocity of sand as it hits belt B - velocity of belt B)

The Attempt at a Solution

V_B=8cos(15°)+8sin(15°) = -7.727[tex]\hat{i}[/tex]+2.071[tex]\hat{j}[/tex](ft/s)

My trouble is with finding V_S. I know that on the horizontal section of belt A, the sand's velocity (V_S) is -6[tex]\hat{i}[/tex]+0[tex]\hat{j}[/tex](ft/s).

When the sand hits the bend where belt A is not horizontal anymore, but starts to bend into belt B, the velocity of the sand changes.

I think I'm stuck on the trig of what the angle is and how the 5ft contributes to finding the correct angle that the sand runs into belt B.

 

[issacnewton]

when the sand leaves the belt A, its in free fall. use projectile motion analysis to find
the velocity 5 ft below, when it hits belt B.