position for this part of the problem) and take +xleftward and +ydownward. The result of part (a) implies v0= Rω= 6.3 m/s, and we see from the figure that (with these positive direction choices) its components are 00cos305.4 m ssin303.1 m s.xyvv=°==The projectile motion equations become xvtyvt gtxy==+212and. We first find the time when y= H= 5.0 m from the second equation (using the quadratic formula, choosing the positive root): 220.74s.yyvvgHtg−++Then we substitute this into the xequation and obtainx5407440...mssm.bgbg9. (a) We find its angular speed as it leaves the roof using conservation of energy. Its initial kinetic energy is Ki= 0 and its initial potential energy is Ui= Mghwhere 6.0sin303.0 mh=(we are using the edge of the roof as our reference level for computing U). Its final kinetic energy (as it leaves the roof) is (Eq. 11-5)
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