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Unformatted text preview: Ph1a Solutions 2 CheFung Chan (cchan@caltech.edu), Fall 2007 Each homework problem is worth 5 points. Please disregard the point values listed on the problem itself. Use these instead. 2.1 QP2 (5 points) 2.1.a (1 point) The trajectory of the boulder is given by x ( t ) = v t cos y ( t ) = v t sin  1 2 g t 2 The constraint that the boulder reaches its maximum height at 200 s implies that this is the point where the yvelocity is zero. By symmetry, this must also occur at half of the horizontal distance. x ( t = 200 s) = 50000 m v y (200 s) = dy dt (200 s) = First we use the x constraint to solve for . x (200 s) = (1000 m s 1 )(cos )(200 s) = 50000 m cos = 50000 m 1000 m s 1 200 s = 1 4 = arccos(1 / 4) 1 . 32 rad or 75 . 5 Next, we calculate the derivate of y ( t ) and apply the velocity constraint to find g . y ( t ) = v sin  g t y (200 s) = (1000 m s 1 ) sin(1 . 32) (200 s) g = 0 g = (1000 m s 1 ) sin(1 . 32) 200 s 4 . 84 m s 2 2.1.b (1 point) Simply plug t = 200 s into the trajectory equations: y (200 s) = (1000 m s 1 ) sin(1 . 32) (200 s) . 5 (4 . 84 m s 2 ) (200 s 2 ) 96800 m 2.1.c (1 point) We need to find t such that y ( t ) = 10000 m. Again, substitute into the trajectory equation and solve the resulting quadratic equation for t . y ( t ) = v t sin  1 2 g t 2 = 10000 m 1 2 g t 2 v t sin + 10000 m = 0 t = v sin radicalBig ( v sin ) 2 4( 1 2 g )(10000 m) 2( 1 2 g ) 390 . s or 10 . 6 s . Were interested in the time when the boulder is coming down, so t = 390 . s. 1 2.1.d (1 point) To get the ships velocity, we integrate the acceleration with respect to time....
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This note was uploaded on 06/05/2008 for the course PH 1a taught by Professor Goodstein during the Fall '07 term at Caltech.
 Fall '07
 Goodstein
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