Posted by Kali on Wednesday, September 12, 2012 at 3:45pm.
the potential energy at the top of the leap is equal to the (vertical) kinetic energy at the start
m * g * h = .5 * m * [v * sin(45º)]^2
the m's cancel
(2 * g * h) / [sin(45º)]^2 = v^2
g = 9.8 m/s^2
145 = v^2
What minimum speed does a 170 puck need to make it to the top of a 3.5 -long, 23 frictionless ramp?
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