hw3solution

# hw3solution - P1112 Fall 2008 Homework#3 Solutions 1 We...

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P1112 Fall 2008 Homework #3 Solutions 1 . We define our coordinate system such that at t=0, x=y=0, and y is positive in the upward direction. (a) To find the time we only need to consider the vertical motion. The time to fall from a height h=0.75m is t= 2 h |a| = 2 x0.75 m 9.8m/s 2 =0.391 s (b) The horizontal velocity does not change during the motion so v x = d t = 1.40 m 0.391 s =3.58m/s (c) The vertical velocity right before impact is v y =a t = -9.8 m/s 2 x 0.391 s=3.83 m/s The magnitude of the total velocity is then |v|= v x 2 +v y 2 = 5.24 m/s The direction is θ =arctan |v y | |v x | =47 ° down from horizontal 2. Subscript definitions: b stands for boat, r stands for river, and s stands for shore. m/s 0 . 6 b,r = v . i v ˆ ) m/s 0 . 2 ( s r, = r . (a) s r, b,r s b, v v v r r r + = . To maximize s b, v , point the boat downstream ( x + direction) so that r b, v r points in the same direction as s r, v r . + x + y w = 100 m m/s 0 . 2 s r, = v r b, v r s r, v r s b, v r

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m/s 0 . 8 m/s 0 . 2 m/s 0 . 6 s,max b, = + = v . To minimize s b, v , point the boat upstream ( x ! direction) so that r b, v r points in the opposite direction to s r, v r . m/s 0 . 4 m/s 0 . 2 m/s 0 . 6 min s, b, = ! = v . (b) Now j v ˆ ) m/s 0 . 6 ( r b, = r . i j v v v ˆ ) m/s 0 . 2 ( ˆ ) m/s 0 . 6 ( s r, b,r s b, + = + = r r r . m/s 3 . 6 ) m/s 0 . 6 ( ) m/s 0 . 2 ( 2 2 s b, = + = v . ° = !
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hw3solution - P1112 Fall 2008 Homework#3 Solutions 1 We...

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