Pre-Calc Homework Solutions 415

Pre-Calc Homework Solutions 415 - Section 10.4 Quick Review...

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Unformatted text preview: Section 10.4 Quick Review 10.4 1. 50 cos 25 , 50 sin 25 2. 80 cos 120 , 80 sin 120 45.315, 21.131 40, 40 3 2. Use R 3. To find the x-intercepts, solve 2x the quadratic formula: x 11 2 415 Section 10.4 Exercises 1. Solve vxt (840 cos 60)t 2 21,000 for t: t v02 9.8 50 seconds. v0 sin 2 ; solve 24,500 g sin 90 11x 40 0 using 112 4(2)( 40) 2(2) for v0: v0 3. (a) t 490 m/sec. 2(500)sin 45 9.8 5002 sin 90 9.8 2v0 sin g v0 g sin 2 2 72.154 seconds; 25,510 m 5 or 2 5 8. The x-intercepts are , 0 and ( 8, 0). For the 2 R y-intercept, find f (0) y-intercept is (0, 40). 2(0)2 11(0) 40 40. The (b) y 25.510 km downrange g 2v02 cos2 4. At the vertex, f (x) vertex is 11 , 4 11 ,2 4 441 . 8 4x 11 2 4 11 11 0 and x 11 4 11 . Then the 4 x2 (tan ) x 4020 m 40 (c) ymax 9.8 50002 + (tan 45)5000 2(500)2 cos2 45 (v0 sin )2 2g (500 sin 45)2 2(9.8) 6377.551 m 5. To find the x-intercepts, solve 20x x 2 0: x 0 or 20. The x-intercepts are (0, 0) and (20, 0). For the y-intercept, find y (0): it is already known to be 0. So the y-intercept is (0, 0). 6. At the vertex, g (x) 20 2x 0 and x vertex is (10, 20(10) 102) (10, 100). 7. y y 8. y t cos x cos x 2 4. With the origin at the launch point (so the ground is t y (v0 sin ) t 32 2 t t 2 when y 1 2 gt . 2 32), use 10. Then the C 2, so C. y 2. 2 cos 2 C (32 sin 30)t t2 1 (32)t 2 2 2 seconds (v0 cos ) t (32 cos 30)2 32 3 55.426 C1 and y ( 1) 2 1 3 t 3 C1t C1 C2 C2 4, so C1 10 3 Then x 3. 5, so 5. Use y 16t t 2 y ( 1) y( 1) C2 y 9. 25 3 C1 1 feet away (horizontally). (v0 sin )t 22 16 1 2 gt 2 1 ( 1)3 3 25 3 3( 1) C2 6.5. t3 3 2t 6.5 346 0 2.135 seconds (by the quadratic (v0 cos ) t 3t dt y ke 16 t 11 2 dy 16 y formula). Substitute that into x t C (44 sin 45)t to obtain x stopboard. 20 so k 4 ln 16 16 y(0) y 10. 16 dy 4 2y 1 ln 4 2 66.4206. 66.421 feet from the y k 4e t 6. With the origin at the launch point, use y (v0 sin )t (44 sin 40 ) 1 2 gt . 6.5 (44 sin 40 )t 16t 2 2 (44 sin 40 )2 416 1.974 sec 32 x dx 2y ke x2 t 1 2 x 2 C At t 1.974, x (44 cos 40 )t 66.5193 ft 0.0987 feet 1.18 inches 4 y y(0) y 2y 2 Thus the shot would have gone farther. 2 k x2 e 2 k 2 1 so k 2 2 e x 2 ...
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