by barnamah » Thu 05 May, 2011 06:46
Because the sand moves at 15° angle, we have x component and y component. As we want to find the distance between the conveyer belt and the bottom of the pipe which is located 3m bellow the canveyer belt, we need to find the horizontal distance d which is the x distance.
we have:
Initial velocity 6.0m/s
`V_0=6.0m/s`
setting the center of the conveyer belt at 0 in our x and y axis the( please view the image)
x initial `x_i=0m`
x final `x_f`=? we want to find
y initial `y_i=3m`
y final `y_f=0m`
the angle 15° `theta=15°`
1) we find the horizontal velocity `v_(ix)=v*costheta=6.0*cos(15)=5.8m/s`
2)We find vertical velocity `v_(iy)=v*sintheta=6.0*sin(15)=1.6m/s` and velocity due to gravity 8.9m/`s^2` is added to this velocity. Using formula
`y_f=y_iv_(iy)*Δt1/2a_yΔt^2` EQ1
`y_f=0m`
`y_i=3m`
`v_(iy)=1.6m/s`
`a_y=9.8m/s^2`
substitute the above values in EQ1 we have:
`0=31.6*Δt1/2(9.8)Δt^2`
`0=31.6*Δt4.9Δt^2`
This is now a quadratic equation and it is the same as `4.9x^21.6x+3=0` and solving for x. So we solve for t.
Solving for t we get t=0.96 and t=0.63 out of 2 values the 0.96 is not valid because of the negative time so we use t=0.63s
We can use vertical velocity (velocity in y direction) to find out how long it took to reach to the ground.
`Δt=0.63s` so it takes the sand to reach to the ground in 0.63s. So this is the same time for x direction for the sand to travel the distance d which we call it `x_f`
Using this value we can use `x_f=x_i+v_(ix)*Δt` we can find the `x_f`
`x_f=x_i+v_(ix)*Δt`
`x_f=0+5.8*0.63`
`x_f=3.654m`
`:.` the distance d is 3.654m
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