warmupsolsp08 - January11,2008 PHY2053Discussion...

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January 11, 2008 PHY2053 Discussion Name:                                   UFID: **1. (5 pts) A tower having a height of 100. m is located near a building. An observer  whose eyes are 1.60 m high is on the top of the building. He/she observes the top of the  tower at an angle of 45.0º above the horizontal and the base of the tower at an angle  30.0º below the horizontal. What is the height of the building the observer is on? Let x be the distance between the building and the tower, y be the height of the part of the   tower form ground to the observer’s eye level, and y’ be the height of the rest of the tower.  Using the definition of tan , we get θ tan30º = y/x (1) and tan45º = y’/x (2) Since y+y’ = 100, we get 100 = x(tan30º+tan45º)   x = 100/(tan30º+tan45º) = 63.4m 
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This note was uploaded on 12/13/2011 for the course PHY 2053 taught by Professor Buchler during the Fall '06 term at University of Florida.

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warmupsolsp08 - January11,2008 PHY2053Discussion...

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