How far does a block go up a frictionless ramp?

AI Thread Summary
A block with an initial velocity of 3.00 m/s is sliding up a frictionless incline at a 22.0° angle. The initial calculations for the block's vertical velocity and distance traveled were incorrect, leading to confusion about the correct distance up the incline. To find the accurate distance, it's essential to consider the component of gravitational acceleration acting along the incline. Using energy conservation principles, the initial kinetic energy can be equated to the gravitational potential energy at the highest point to determine the height and distance along the incline. Correctly applying these concepts will yield the accurate distance the block travels before coming to rest.
Talby
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A block is given an initial velocity of 3.00 m/s up a frictionless 22.0° incline. How far up the incline does the block slide before coming (momentarily) to rest? So basically, how far does it go up before the block starts falling.
http://img11.paintedover.com/uploads/11/p522.gif

At first glance, this seemed really simple to me. Here's what I tried:

The initial velocity in the Y direction would be 3sin(22) = 1.12 m/s

Then I used the formula
Vf2 = Vi2 + 2*(-9.8)*(delta Y)
-->0 = 1.26 - 19.6(delta Y)
so delta Y = .064 m.

But that is straight up, and I need it along the incline. So, I did

sin(22) = (.064)/x
and x gives me .17 as the distance up the incline.

However, neither .064m or .17m is correct. Can anyone tell me what I'm doing wrong? Am I totally off, or just missing a decimal or something?
 
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Use that same kinematic equation (v_f^2 = v_i^2 + 2ax) but use the components along the incline. Hint: Figure out the component of the acceleration due to gravity along the incline.
 
surely you can do it by energy considerations - you know the KE at the start, and no GPE, then at the top there will be no KE momentarily, so you can work out h?
 
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