87_Physics ProblemsTechnical Physics

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

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 .

Ask a homework question - tutors are online