Ch04-p099 - 99 We choose horizontal x and vertical y axes such that both components of v0 are positive Positive angles are counterclockwise from x

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() 00 ˆˆ cos i sin j. vv v g t θθ =+ G (a) With v 0 = 30 m/s and θ 0 = 60°, we obtain (15i+6.4 j) m/s v = G , for t = 2.0 s. The magnitude of v G is 22 | | (15 m/s) (6.4 m/s) 16 m/s. v = G (b) The direction of v G is 1 tan [(6.4m/s)/(15m/s)] 23 , == ° measured counterclockwise from + x . (c) Since the angle is positive, it is above the horizontal. (d) With t = 5.0 s, we find (15i 23j) m/s v =− G , which yields | | (15m/s) ( 23m/s) 27 m/s. v = G (e) The direction of v G is 1 tan [( 23m/s)/(15m/s)] 57 = ° , or 57
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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.

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