Chapter 487P4.13(a)The time of flight of the first snowball is the nonzero root of yyvtatfiyiy=++1121200 25070012980225070047921° −=°=.sin..(.)sin....ms±mss22bgbgejtttThe distance to your target isxxvtfixi−==°=125 070 0 4 7941 0.cos...smaf.Now the second snowball we describe byvtatiy+22212025049022=−in.2θt510=insti241 025 05 1012822..cos.sinsincosm mssm==bg afafθθ0321incos=Using sinsin cos=we can solve 1222in=2064321=−sin.and 2=°20 0(b)The second snowball is in the air for time t5 105 10201 75°=inin.s s, so youthrow it after the first by
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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 .