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A box moves down a conveyor belt the is angled at 15 degrees, at a rate of 6m/s. The box then enters a dump off that is 3m belo...Asked by Confused
A box moves down a conveyor belt the is angled at 15 degrees, at a rate of 6m/s. The box then enters a dump off that is 3m below the end of the belt, what is the horizontal distance between the dump off and the belt?
I have no idea where to start with this...I know you have x and y components of something but I don't know what to do...any help in the right direction would be appreciated.
I have no idea where to start with this...I know you have x and y components of something but I don't know what to do...any help in the right direction would be appreciated.
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Answered by
bobpursley
:You have to figure the time it took to fall 3m. It had an initial vertical velocity downward of -6sin15.
h=vi*t -4.9t^2
solve for t.
Then, you know the intial horizontal velocity is 6cos15, and the time, so what distance did it travel horizontally?
h=vi*t -4.9t^2
solve for t.
Then, you know the intial horizontal velocity is 6cos15, and the time, so what distance did it travel horizontally?
Answered by
Confused
You can't solve for t from that though can you? If you use quadratic equation its a negative sign under the root and you can't take the root of that. I'm confused:S
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