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Unformatted text preview: x h y v O L 86. If v PA is the velocity of the airplane with respect to the air, v PG the velocity of the airplane with respect to the ground, and v AG the velocity of the air(wind) with respect to the ground, then v PG = v PA + v AG , as shown in the diagram. From the diagram we find the two components of v PG : v PGW = v AG sin 45 = (100 km/h) sin 45 = 70.7 km/h; v PGN = v PA v AG cos 45 = 200 km/h (100 km/h) cos 45 = 129.3 km/h. For the magnitude we have v PG = ( v PGW 2 + v PGN 2 ) 1/2 = [(70.7 km/h) 2 + (129.3 km/h) 2 ] 1/2 = 147 km/h . We find the angle from tan = v PGW / v PGN = (70.7 km/h)/(129.3 km/h) = 0.547, which gives = 28.7 west of north . 87. If v AG is the velocity of the automobile with respect to the ground, v HG the velocity of the helicopter with respect to the ground, and...
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This note was uploaded on 03/08/2010 for the course PHYSICS 7A/7B taught by Professor All during the Fall '08 term at University of California, Berkeley.
- Fall '08