physics
posted by cynthia .
a bicyclist is coasting on level ground at 32.35m/s. The bicyclist coasts up an incline of 20.5m at a 30degree angle. How fast is the bicyclist traveling when he grabs the limb?

What "limb" are you talking about?
If he has a frictionless bike, his kinetic energy will be reduced by the work done against gravity.
(M/2) delta (V^2) = 20.5 sin 30 * M g
Delta (V^2) = 41 sin 30 * g 
by limb i mean a limb on a tree (srry about that).. and when i set Eg and Ek equal to each other i get 14m/s but the answer is 29m/s

41 sin 30*g = 200 m^2/s^2. The original V^2/2 was 1039 m^2/s^2. After coasting uphill, V^2/2 = 1039  200 = 839 m^2/s^2, and so V = 29.0 m/s
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