64. We first use Eq. 4-26 to solve for the launch speed of the shot: 202(tan ).2( cos )gxyyxvθ−=−′With 34.10 ,=°02.11 my=and ( , )(15.90 m,0)xy=, we find the launch speed to be 11.85 m/s.v′ =During this phase, the acceleration is 2220(11.85 m/s)(2.50 m/s)40.63 m/s .(1.65m)vvaL′ −−===Since the acceleration along the slanted path depends on only the force components along
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.