87_Physics ProblemsTechnical Physics

# 87_Physics ProblemsTechnical Physics - 88 P4.15 Motion in...

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Unformatted text preview: 88 P4.15 Motion in Two Dimensions h= so b g vi2 sin 2θ i vi2 sin 2 θ i ; R= ; 3h = R , g 2g b 2 3 vi2 sin 2 θ i vi sin 2θ i = g 2g g 2 sin 2 θ i tan θ i = = 3 sin 2θ i 2 4 = 53.1° . thus θ i = tan −1 3 or FG IJ HK *P4.16 (a) To identify the maximum height we let i be the launch point and f be the highest point: d i + 2b− g gb y 2 2 v yf = v yi + 2 a y y f − yi 0= y max = vi2 vi2 2 sin θ i max −0 g 2 sin θ i . 2g To identify the range we let i be the launch and f be the impact point; where t is not zero: 1 ay t 2 2 1 0 = 0 + vi sin θ i t + − g t 2 2 2 vi sin θ i t= g y f = yi + v yi t + bg 1 axt 2 2 2 v sin θ i + 0. d = 0 + vi cos θ i i g x f = xi + v xi t + For this rock, d = y max vi2 sin 2 θ i 2 vi2 sin θ i cos θ i = g 2g sin θ i = tan θ i = 4 cos θ i θ i = 76.0° (b) Since g divides out, the answer is the same on every planet. (c) The maximum range is attained for θ i = 45° : d max vi cos 45° 2 vi sin 45° g = = 2.125 . d gvi cos 76° 2 vi sin 76° So d max = 17d . 8 ...
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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