Is This Equation Correct for Horizontal Projection?

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SUMMARY

The discussion centers on deriving the relationship between vertical distance fallen (y) and horizontal distance traveled (X) for a block sliding off a horizontal surface with an initial speed (Vo). Participants confirm that using the horizontal projection equations is appropriate, leading to the expression y = 0.5g(x/Vo)^2. This equation can be simplified to (square root of y/4.9) = X/Vo, accurately representing the physics of horizontal projection. The consensus is that the original form of the equation is sufficient and should not be overly simplified.

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Physicsisfun2005
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I think this physics problem deals only with horizontal projection...let me know what you think:

A block slides off a horizontal surface with speed Vo. Write an expression relating y, distance fallen, to X, the horizontal distance traveled.
i think i can use the horizontal proj. equations:

R=volt (I think you can substitute X for R)

h=.5gt^2 (I think you can substitute y for h)

I get a combined equation of y=.5g(x/Vo)^2..which simplfied b/c .5g=4.9
Y/4.9=(X/Vo)^2...which i think can simplify to the
(square root of y/4.9)=X/Vo
 
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Originally posted by Physicsisfun2005
I get a combined equation of y=.5g(x/Vo)^2
That is correct, and I would leave it in exactly that form. The rest of your "simplifications" really just seemed to make a mess of it. You are asked to express y as a function of x (or x as a function of y), and that's exactly what you've done. You're finished!

- Warren
 


Yes, you are correct. The problem only deals with horizontal projection, as the block is sliding off a horizontal surface. Your approach of using the horizontal projection equations is correct. By substituting X for R and y for h, you can arrive at the expression y=.5g(x/Vo)^2, which can be simplified to (square root of y/4.9)=X/Vo. This equation relates the distance fallen, y, to the horizontal distance traveled, X. Great work!
 

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