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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 5 5 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 5 2 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 5 2 1 } 2(116) 2 3 c 2 os 2 45° } 2 x 2 1 (tan 45°) x 5 2 } 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. 13. (a) With the origin at the launch point, use y 5 2 1 } 2 v 0 2 c 3 o 2 s 2 20° } 2 x 2 1 (tan 20°) x . Set x 5 315 and y 5 37 2 3 5 34, then solve to find v 0 5 < 149.307 ft/sec. (b) Solve v 0 (cos 20°) t < 149.307(cos 20°) t 5 315 to find t < 2.245 seconds. 14. In the formula for range, sin 2 a 5 sin 2(90 2 a ). 15. Use R 5 } sin 2 a : sin 2 a 5 } (9.8) 4 ( 0 1 0 6 2 ,000) } 5 0.98; a 5 } sin 2 1 2 0.98 } < 39.261 8 or a 5 90 2 } sin 2 1 2 0.98 } < 50.739 8 . 16. (a) Substitute 2 v 0 for v 0 in the formula for range. (b) To increase the range (and height) by a factor of 2, increase v 0 by a factor of ˇ 2 w < 1.41. That is an increase of < 41%. 17. With the origin at the launch point, y 5 2 1 } 2 v 0 2 c 3 o 2 s 2 40° } 2 x 2 1 (tan 40°) x . Setting x 5 73 } 5 6 } and y 5 2 6.5 and solving for v 0 yields v 0 < 46.597 ft/sec. 18. y ( t ) 5 v 0 (sin a ) t 2 } 1 2 } gt 2 , and we know the maximum height is } ( v 0 s 2 in g a ) 2 } and it occurs when t 5 } v 0 s g in a } . Substituting t 5 } v 0 2 si g n a } into the equation for y ( t ) gives a height of } 3( v 0 8 si g n a ) 2 } , which is three-fourths of the maximum height. 19. Integrating, } d d t } r ( t ) 5 c 1 i 1 ( 2 gt 1 c 2 ) j . The initial condition on the velocity gives c 1 5 v 0 cos a and c 2 5 v 0 sin a . Integrating again, r ( t ) 5 (( v 0 cos a ) t 1 c 3 ) i 1 1 } 1 2 } gt 2 1 ( v 0 sin a ) t 1 c 4 2 j . The initial condition on the position gives c 3 5 x 0 and c 4 5 y 0 . 20. With the origin at the launch point, y max 5 68 ft. Then v 0 5 } ˇ s 2 w i y n m w a ax w g w } < v 0 5 < 79.107 ft/sec. 21. The horizontal distance is 30 yd 2 6 ft 5 84 ft. Then 84 5 ( v 0 cos a ) t , where a 5 tan 2 1 1 } 6 4 8 5 } 2 < 56.5° and v 0 5 } 1 s 6 i ˇ n a 1 w 7 w } (from Exercise 20). So t 5 } v 0 c 8 o 4 s a } 5 5 = } 15 1 ˇ 19 1 w 7 w } < 1.924 seconds. Then y 5 ( v 0 sin a ) t 2 } 1 2 } gt 2 5 (16 ˇ 1 w 7 w ) 1 } 15 1 ˇ 19 1 w 7 w } 2 2 } 1 2 } (32) 1 } 15 1 ˇ 19 1 w 7 w } 2 2 < 67.698 ft. The height above the ground is 6 ft more than that, < 73.698, and the height above the rim is about 73.698 2 70 5 3.698 feet. 22. The projectile rises straight up and then falls straight down, returning to the firing point. 21 1 } 6 4 8 5 } 2 } 4 ˇ 1 w 7 w 84 tan a } 16 ˇ 1 w 7 w ˇ 2 w (6 w 8 w )( w 3 w 2 w ) w }} sin 56.505° v 0 2 } g 1260 }}} cos 20° ˇ 3 w 1 w 5 w t w an w 2 w 0 w ° w 2 w 3 w 4 w 1 1 !

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