03_InstSolManual_PDF_Part9

03_InstSolManual_PDF_Part9 - 5 17.7 m / s v y 5 v y 1 a y t...

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(b) Use the horizontal motion to find the time in the air. The grasshopper travels horizontally gives Find the vertical displacement of the grasshopper at The height of the cliff is 4.66 m. 3.23. Set Up: Let be upward. Solve: (a) gives and (b) At the highest point and At all points in the motion, downward. (c) Find when and Reflect: The acceleration is the same at all points of the motion. It takes the water to reach its maximum height. When the water reaches the building it has passed its maximum height and its vertical component of velocity is downward. 3.24. Set Up: Example 3.5 derives Solve: The maximum range occurs when and 3.25. Set Up: Example 3.5 gives and Solve: (a) R is a maximum when (b) (c) 3.26. Set Up: Example 3.5 derives Solve: is proportional to so when is tripled, is increased by a factor of 9. 13.5 km. R max 5 9 1 1500 m 2 5 1.35 3 10 4 m 5 R max v 0 v 0 2 R max R max 5 v 0 2 g . R 5 v 0 2 sin 2 u 0 g . h 5 v 0 2 sin 2 45° 2 g 5 v 0 2 4 g 5 R max 4 5 5.75 m R max 5 v 0 2 g 5 1 15.0 m / s 2 2 9.80 m / s 2 5 23.0 m u 0 5 45°. h 5 v 0 2 sin 2 u 0 2 g . R 5 v 0 2 sin 2 u 0 g u 0 5 45°. 2 u 0 5 90° sin 2 u 0 5 1, R 5 v 0 2 sin 2 u 0 g . 2.04 s t 52 v 0 y a y 52 20.0 m / s 2 9.80 m / s 2 5 v 5 " v x 2 1 v y 2 5 "1 15.0 m / s 2 2 1 1 2 9.41 m / s 2 2
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Unformatted text preview: 5 17.7 m / s v y 5 v y 1 a y t 5 1 25.0 m / s 21 sin 53.1° 2 2 1 9.80 m / s 2 21 3.00 s 2 5 2 9.41 m / s, v x 5 v x 5 15.0 m / s, y 2 y 5 v y t 1 1 2 a y t 2 5 1 25.0 m / s 21 sin 53.1° 21 3.00 s 2 1 1 2 1 2 9.80 m / s 2 21 3.00 s 2 2 5 15.9 m t 5 3.00 s: y 2 y a 5 9.80 m / s 2 v 5 " v x 2 1 v y 2 5 15.0 m / s. v y 5 v x 5 v x 5 1 25.0 m / s 2 cos 53.1° 5 15.0 m / s, u 5 53.1° cos u 5 45.0 m 1 25.0 m / s 21 3.00 s 2 5 0.600; x 2 x 5 v 1 cos u 2 t x 2 x 5 v x t 1 1 2 a x t 2 v y 5 v sin u . v x 5 v cos u , a y 5 2 9.80 m / s 2 . a x 5 0, 1 y y 2 y 5 v y t 1 1 2 a y t 2 5 1 1.15 m / s 21 1.10 s 2 1 1 2 1 2 9.80 m / s 2 21 1.10 s 2 2 5 2 4.66 m t 5 1.10 s: t 5 x 2 x v x 5 x 2 x v cos 50.0° 5 1.10 s x 2 x 5 v x t 1 1 2 a x t 2 x 2 x 5 1.06 m. Motion in a Plane 3-9...
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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