Question

a. 3s b. 4s c. 8s d. 16s 28. How far did the shell travel horizontally when it hit the ground? Assume there were no obstacles in the way and the landing point is at the same elevation as the launch point a. 1532m b. 885m c. 750m d. 375m c 68" x 23.0 m A 49° 29. Find the magnitude of the resultant vector R: R 2A-B+ .2C

29 and 30

Answer 1

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The options for question 30 are wrong. The answer is coming as 23.15 m and no option is close to that value. I have double checked the solution. I have also used vector calculator which gives same value as mine. Therefore, there is something wrong the question 30. See for yourself.

The method is perfectly fine.  

The options for question 30 are wrong. The answer is coming as 23.15 m and no option is close to that value. I have double checked the solution. I have also used vector calculator which gives same value as mine. Therefore, there is something wrong the question 30. See for yourself. The method is perfectly fine. R = 2A - B + 0.2 C R_x = 2 A_x - B_x + 0.2 C_x R_x = 2(25 times cos 49 degree) - 23 cos 15 degree + 0.2632 times cos 68 degree R_x = 12.984 m R_y = 2(25 times sin 49 degree) - 23 sin 15 degree - 0.2(32 times sin 68 degree 0 R_y = 25.848 m therefore Vertical-bar R Vertical-bar = squareroot R_x^2 + R_y^2 Vertical-bar R Vertical-bar = 28.9 m \r\n R = 0.2A + 0.5B - 0.5 C R_x = 0.2 times 0.5 times cos 49 degree + 0.5 times 23 times cos 15 degree - 0.5(32 times cos 68 degree) R_x = 8.3947 m R_y = 0.2 times 25 times sin 49 degree + 0.5 times 23