Is This Equation Correct for Horizontal Projection?

AI Thread Summary
The discussion confirms that the physics problem focuses on horizontal projection as a block slides off a horizontal surface. The correct expression relating the vertical distance fallen (y) to the horizontal distance traveled (X) is derived as y = 0.5g(x/Vo)^2. Participants agree that this formulation is appropriate and that further simplifications complicate the equation unnecessarily. The final expression effectively relates y and X, demonstrating a clear understanding of the problem. The consensus is that the initial approach and derived equation are correct and sufficient for the problem at hand.
Physicsisfun2005
Messages
70
Reaction score
0
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=VoT (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
 
Physics news on Phys.org
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!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top