posted by Anonymous .
A coin is tossed into the air with an initial speed of 8.00 m/s. In the absence of air resistance, how high does the coin go above its point of release
It rises a distance h such that the potential energy gain (m g h) equals the initial kinetic energy loss ((1/2) m v^2 ). Therefore v^2 = 2 g h
v is the initial speed. It is assumed that the coin is tossed straight up, otherwise the answer is different
Solve for h