Max velocity up a frictionless inclined plane

[olliepower]

Homework Statement

The air table is approximately 25 cm square. If the incline has a rise of 1cm over that distance, what is the maximum velocity up the slope that a puck may have and not hit the upper wall

Homework Equations

ymax = (V^2sin(θ))/(2g)
xmax = 2V^2sin(θ)cos(θ)/g

The Attempt at a Solution

0.25 = (2 v^2 sin(2.29) cos(2.29))/9.81

V = 5.54

then I took my result substituted it into the y max equation.

ymax = ((5.54^2)*sin(2.29))/(2*9.81) = .06 M

Does this mean that if the puck was launched at that velocity, it would go off the end of the table? Should first be setting ymax = .01? If I do that, it does not seem that the puck would go anywhere in the x direction.

 

[issacnewton]

I think you mean the minimum velocity the puck should have. Because any velocity beyond the minimum velocity would be enough for the puck to hit the top portion of the table..