9.4.15 Question Help The acceleration (a vector quantity) of gravity on a sky di

Question

9.4.15 Question Help The acceleration (a vector quantity) of gravity on a sky diver is 9.8m/s2 If the force of the wind also causes an acceleration of 0.6 m/s at an angle of 12° 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 Magnitude Standard-position Angle Enter your answer click Check Answer part

Explanation / Answer

S = 9.8 m/s2 (270o)

W = 0.6 m/s2 (12o)

F = S + W = 9.8(cos270 i + sin270 j) + 0.6(cos12 i + sin12 j) = 0.6cos12 i - (9.8 - 0.6sin12) j = 0.587 i - 9.675 j m/s2

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