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
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.