When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and thus slows him down. Suppose the weight of the sky diver is 905 N and the drag force has a magnitude of 1005 N. The mass of the sky diver is 92.3 kg. Take upward to be the positive direction. What is his acceleration, including sign?

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11 years ago

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ma=-mg +F
a= F/m- g = 1005/92.3 - 9.8 =
=10.9 -9.8 = 1.1 m/s²

answered by Elena

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Question

When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and thus slows him down. Suppose the weight of the sky diver is 905 N and the drag force has a magnitude of 1005 N. The mass of the sky diver is 92.3 kg. Take upward to be the positive direction. What is his acceleration, including sign?

1 answer

ma=-mg +F
a= F/m- g = 1005/92.3 - 9.8 =
=10.9 -9.8 = 1.1 m/s²

Elena · Answer

ma=-mg +F a= F/m- g = 1005/92.3 - 9.8 = =10.9 -9.8 = 1.1 m/s²