If something needs to fall 4 feet to the ground and 3 feet away from its falling point, how fast must the material go if the conveyor belt is a flat plane or 0 degrees. I came up with 5.9 ft per sec but I think it might be a little fast. Can someone show me where I might have gone wrong or am I pretty close?

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the formula is \[v=\sqrt{2gh}\] here g=9.8m/s^2=32.152231ft/s^2 and h=3ft so ... velocity = 13.8893 ft/s

d = Vt - (1/2) a t^2 V in the y direction equals 0 so the Vt term equals 0 d = h = 4 a =32.2 ft/s^2 so \[t = \sqrt{2h/g}\] = .498 s After you find the time that the object is in the air, you can then solve for the velocity in the x direction. It will remain constant because there is no acceleration in the x-direction so d = vt so d in this case would be the x distance which is 3 v = d/t = 3/.498 = 6.02 ft/s You were pretty close

The velocity term in the first equation I wrote is the initial velocity in the y-direction. That is what is equal to zero. This is because the conveyor is at 0 degrees, so the object comes off moving solely in the x-direction and accelerating solely in the y-direction. I should have been more clear.

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