2220011(sin)(25.0 m/s)sin 40.0 (1.15 s)(9.80 m/s )(1.15 s)12.0 m.yvtgtθΔ=−=°−=(b) The horizontal component of the velocity when it strikes the wall does not change from its initial value: vx= v0cos 40.0° = 19.2 m/s. (c) The vertical component becomes (using Eq. 4-23) 2sin(25.0 m/s) sin 40.0(9.80 m/s )(1.15 s) 4.80 m/s.yvvgt=−=°−=(d) Since vy> 0 when the ball hits the wall, it has not reached the highest point yet. 38. We adopt the positive direction choices used in the textbook so that equations such as
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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.