Pre-Calc Homework Solutions 416

# Pre-Calc Homework Solutions 416 - 416 Section 10.4 v02 sin...

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7. (a) Use R 5 sin 2 a ; solve 10 5 } 9 v 0 .8 2 } sin 90° for v 0 : v 0 5 7 ˇ 2 w < 9.899 m/sec. (b) Solve 6 5 } (7 ˇ 9.8 2 w ) 2 } sin 2 a for a : sin 2 a 5 0.6, so 2 a 5 sin 2 1 0.6 < 36.870 8 and a < 18.435 8 or 2 a 5 180° 2 sin 2 1 0.6 < 143.130 and a < 71.565 8 . 8. t 55 8 3 10 2 8 sec. Then y (taking down as positive) is } 1 2 } gt 2 < } 1 2 } (9.8)(8 3 10 2 8 ) 2 5 3.136 3 10 2 14 meters or 3.136 3 10 2 12 cm. 9. R 5 sin 2 a (248.8 yd)(3 ft/yd) 5 } 32 f v t 0 / 2 sec 2 } sin 18° v 0 < 278.016 ft/sec or < 189.556 mph. 10. R 5 } sin 2 a 200 5 sin 2 a sin 2 a 5 0.9. Taking the smaller of the two possible angles, a 5 } 1 2 } sin 2 1 0.9 < 32.079 8 . Then y max << 31.339, which is well below the ceiling height. 11. No. For a 5 30°, v 0 5 90 ft/sec, and x 5 135 ft, y 52 1 } 2 v 0 2 3 c 2 os 2 a } 2 x 2 1 (tan a ) x evaluates to < 29.942 feet above the ground, which is not quite high enough. 12. Use y 1 } 2(116) 2 3 c 2 os 2 45° } 2 x 2 1 (tan 45°) x 52} 8 2 41 } x 2 1 x . Set y 5 45, then solve for x using the quadratic formula and taking the larger of the two values: x 5 < 369.255 ft, which is < 0.255 ft < 3.059 inches beyond the pin.
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