energy at top = (1/2)m v^2 + m g h
= 75 (25/2+9.81*7)
energy at bottom the same but h = 0
so
75 (V^2/2) = 75 (25/2+9.81*7)
(note-- the mass does not matter)
Now if he comes up to the same height on the other side he will be moving up at 5 m/s
(1/2)(25) = 9.81 h
where h is the height above the top of the pipe.
A half pipe can be thought of as half of a cylinder with radius of 7.00m. If a skater with mass 75.0 kg, starts with an initial velocity of 5.00 m/s at the top of the ramp, (a), what is his velocity at the very bottom of the ramp? The skater then approaches the other quarter ramp and is launched vertically into the air. How high up does he go? (Assume the ramp is frictionless)
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