Posted by Elise Brown on Saturday, October 13, 2012 at 7:59pm.
This problem has nothing to do with the second sled or the length of the plane as long as it does not overshoot and fly off the top.
Kinetic energy at bottom = potential energy stopped at top of trajectory.
(1/2) m Vi^2 = m g h = m g (x sin theta)
x = Vi^2 / (2 g sin theta)
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