Sand is discharged at A from a conveyor belt and falls onto the top of a stockpile at B. Knowing that the conveyor belt forms an angle [tex]\alpha[/tex]= 20 degrees with the horizontal, determine the speed V0 of the belt.
yf-yi=Vi,y*t - (1/2)(g)(t2)
The Attempt at a Solution
Hello again guys! I know this problem is very similar to the last one that had the snowballs, the only difference being that there is now an upwards 20 degree angle instead of it being flatly horizontal.
Using the values I could relate Vi to, I found the numerical value of the angle * Vi, and plugged the respective ones into the vertical and horizontal equations. I then found in the horizontal equation, what Vi equaled. After this, I plugged the Vi relation into the vertical equation, solved the time( I got t= 2.428 s ) and plugged the information in. I found the initial velocity to be 13.15 m/s...but the answer in the book says it is 23.8 ft/s.
Would anyone be so kind as to help me out with where I went wrong? Do I need to use the same equation I used for the vertical component as I do for the horizontal? Tieing into that question, does this angle mean that there is an acceleration in the horizontal component as well?
Thanks for any replies!
Sorry it's really crude, but it illustrates everything nicely I believe.