(b) The height of the person when he is directly above the second wheel can be found by solving Eq. 4-24. With the second wheel located at 23 m (23/ 2) m34.5 m,x=+=we have2220022(9.8 m/s )(34.5 m)tan3.0 m (34.5 m)tan532cos2(26.52 m/s) (cos53 )25.9 m.gxyy xvθ−=+°−°=Therefore, the clearance over the second wheel is 25.9 m 18 m7.9 mwyyhΔ= − =−=.(c) The location of the center of the net is given by 22sin 2(26.52 m/s) sin(2 53 )0tan69m.2cos9.8 m/svgxxvg⋅°=− =−¡===44. (a) Using the fact that the person (as the projectile) reaches the maximum height over the middle wheel located at 23 m (23/ 2) m
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