ch11-p009 - 9. (a) We find its angular speed as it leaves...

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position for this part of the problem) and take + x leftward and + y downward. The result of part (a) implies v 0 = R ω = 6.3 m/s, and we see from the figure that (with these positive direction choices) its components are 00 cos30 5.4 m s sin30 3.1 m s. x y vv = = The projectile motion equations become xv t yv t g t xy == + 2 1 2 and . We first find the time when y = H = 5.0 m from the second equation (using the quadratic formula, choosing the positive root): 2 2 0.74s. yy vvg H t g −+ + Then we substitute this into the x equation and obtain x 54 074 40 ... ms s m . bg b g 9. (a) We find its angular speed as it leaves the roof using conservation of energy. Its initial kinetic energy is K i = 0 and its initial potential energy is U i = Mgh where 6.0sin30 3.0 m h = (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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