Question

Sand moves without slipping at 6.0 m/s down a conveyer that is tilted at 15∘. The sand enters a pipe h = 4.5 m below the end of the conveyer belt, as shown in the figure

What is the horizontal distance d between the conveyer belt and the pipe?

Answer #1

- Velocity of sand v=6 m/sec

- Inclination of conveyor θ=15°

- Height of conveyor belt h=4.5m

velocity of sand into a vertical and horizontal component

The horizontal component of velocity of sand

vx=6cos(15°)

vx=5.796 m/sec

The vertical component of velocity of sand

vy=vsin(15°)

vy=6×sin(15)=1.554m/s

Vertical acceleration of sand is,

ay=9.81 m/sec2

quation of motion,

s=ut+12at2s=ut+12at2

Consider motion in vertical direction

h=vyt+12at2

4.5=1.554t+4.905t2

t=0.814sec

Along the horizontal direction, acceleration is zero.

Velocity=DistanceTimed=vx×t=5.796*.814=04.717m

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