Question

The acceleration (a vector quantity) of gravity on a sky diver is 9.8 m/s. If the force of the wind also causes an acceleration of 1.9 m/s- at an angle of 13° above the horizontal, what is the resultant acceleration of the sky diver? Set up the problem by representing each force as a vector. Let S represent the downward force on the sky diver due to gravity, let W represent the upward force on the sky diver due to the wind, and let R represent the resultant vector. Vector Standard-position Angle Magnitude 9.8 m/s2 1.9 m/s2 270 13 m/s4 at an angle of from the vertical The resultant acceleration of the sky diver is 9.6 (Round to one decimal place as needed.)

Answer 1

1.9 mis? n 49.8 m/s2 ş. 9.8m152 (-3) W = 1.9 Los 13 ( [] + 1.9 Sim 13 (3) mis? = 1.85 î + 0.43 I mis? resultant acc. = 3 the = (1.857 +1 0.43 -9.8)]}mis? - (1.85 î + 9.37(-3) ]miş? magnitude = 5(1-85)2 + (9.37)2 mis? = 53.4225 + 87.7969 mis? = 9.55 mls? tano = y. -9.37 = -5.065 n 1.85 = 0 = -78.83 = (360 -78.83) - 281.17 ~ 281

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