A conveyor belt is moving grain into a bin that is 2.30 m below the top of the conveyor belt. The grain does not slip on the conveyor belt that is inclined at 15.0° and they move at a constant speed of 6.00 m/s. In order for the conveyor belt to get the grain into the bin, what must the horizontal distance between the end of the conveyor belt and the bin be?
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- NCSLv 710 months agoFavorite Answer
time to fall is such that (taking down as positive)
h = VsinΘ*t + ½gt²
2.30 m= 6.00m/s*sin15.0º*t + ½*9.8m/s²*t²
2.30 = 1.553t + 4.9t²
for h in meters and t in seconds. This is quadratic with roots at
t = -0.86 s ← ignore negative root
and t = 0.545 s
so horizontal distance
d = Vx*t = 6.00m/s*cos15.0º*0.545s
d = 3.16 m
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