Conveyor belt A, which forms a 20° angle with the horizontal, moves at a constant speed of 4 ft/s and is used to load an airplane. Knowing that a worker tosses duffel bag B with an initial velocity of 2.5 ft/s at an angle of 30° with the horizontal, determine the velocity of the bag relative to the belt as it lands on the belt.
Vi,x = Vi*cos([tex]\theta[/tex])
Vi,y = Vi*sin([tex]\theta[/tex])
(yf-yi) =Vi,y*t - (1/2)*g*t2
rB = rA+rB/A
vB = vA+vB/A
aB = aA+aB/A
The Attempt at a Solution
I think I've gotten the majority of this problem tackled, so I will list what I have done (as I am highly likely to have done something wrong.) As you can see from the picture below, a linear equation (I think?) is required out of the conveyor belt, so that is the first thing I did.
We know that to make a right triangle with the belt, one side is 90, we're given 20, so the final angle is 70. For pure relations' sake to find a slope, I set the bottom length equal to 10, used tan(20)=opp/10, and found the rise/run to be 3.64/10 which came out to be a slope of .364.
I also set my origin at the location of the duffel bag, which was 1.5 feet above the starting location of the conveyor belt line. So, for my linear equation, I found y=.364x-1.5
Then from there, (I think this is right?) I subbed my xf and yf equations into my linear equation. I was then faced with a quadratic, which, when solved for t, I found t=.3209 s. Fast forwarding a bit more after this I found my final x and y distances.
My x final distance was .6947 ft, while my y final distance was -1.2465 ft.
And...this is where I taper off. I am unsure how to proceed next. I know my final distances, the time it took to hit the conveyor belt...but I am not sure how to relate all of this to find the relative velocity of the bag to the belt.
Could anyone spare some help? I would greatly appreciate it!