ch04-p120 - 120. With v0 = 30.0 m/s and R = 20.0 m, Eq....

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Because sin φ = sin (180° – ), there are two roots of the above equation: 1 0 2 sin (0.218) 12.58 and 167.4 . θ == ° ° which correspond to the two possible launch angles that will hit the target (in the absence of air friction and related effects). (a) The smallest angle is 0 = 6.29°. (b) The greatest angle is and 0 = 83.7°. An alternative approach to this problem in terms of Eq. 4-25 (with
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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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