02_InstSolManual_PDF_Part12

# 02_InstSolManual_PDF_Part12 - 2-12 Chapter 2 Reect The...

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Reflect: The answers can be checked several ways. For example, and in gives which agrees with the height calculated in (a). 2.52. Set Up: Take to be downward. and let Solve: (a) gives (b) 2.53. Set Up: Take upward. The initial velocity of the sandbag equals the velocity of the balloon, so When the balloon reaches the ground, At its maximum height the sandbag has Solve: (a) The sandbag is 40.9 m above the ground. The sandbag is 40.1 m above the ground. (b) gives and t must be positive, so (c) (d) gives The maximum height is 41.3 m above the ground. (e) The graphs of and y versus t are given in Figure 2.53. Take at the ground. Figure 2.53 2.54. Set Up: Take to be downward. The acceleration is the slope of the versus t graph. Solve: (a) Since is downward, it is positive and equal to the speed The versus t graph has slope a y 5 30.0 m / s 2.0 s 5 15 m / s 2 . v v . v y v y 1 y a y v y y t t O t O O y 5 0 v y , a y , y 2 y 0 5 v y 2 2 v 0 y 2 2 a y 5 0 2 1 5.00 m / s 2 2 2 1 2 9.80 m / s 2 2 5 1.28 m. v y 2 5 v 0 y 2 1 2 a y 1 y 2 y 0 2 v y 5 0. a y 5 2 9.80 m / s 2 , v 0 y 5 5.00 m / s, v y 5 v 0 y 1 a y t 5 1 5.00 m / s 1 1 2 9.80 m / s 2 21 3.41 s 2 5 2 28.4 m / s t 5 3.41 s. t 5 1 9.80 1 5.00 6 " 1 2 5.00 2 2 2 4 1 4.90 21 2 40.0 2 2 s 5 1 0.51 6 2.90 2 s. 1 4.90 m / s 2 2 t 2 2 1 5.00 m / s 2 t 2 40.0 m 5 0 1 4.90 m / s 2 2 t 2 . 2 40.0 m 5 1 5.00 m / s 2 t 2 y 2 y 0 5 v 0 y t 1 1 2 a y t 2 a y 5 2 9.80 m /
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• Spring '07
• Shoberg
• Velocity, Equals sign, 1 4.90 m, 2 40.0 m, 5 1 5.00 m, 5 15.8 m

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