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Masteriza [31]
3 years ago
5

When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the s

ky diver and, thus, slows him down. Suppose the weight of the sky diver is 968 N and the drag force has a magnitude of 1085 N. The mass of the sky diver is 98.8 kg. Take upward to be the positive direction. What is his acceleration, including sign
Physics
1 answer:
skad [1K]3 years ago
7 0

Answer:

The value a = 1.1842 \ m/s^2

Explanation:

From the question we are told that

The weight of the sky diver is W =  968 \  N

The magnitude of the drag force is D_f  =  1085 \  N

The mass of the sky diver is m  =  98.8 \ kg

Generally the net force acting on the sky diver is mathematically represented as

F = m * a  =  D_f  - W

Hence the acceleration of the sky diver is mathematically represented as

a =  \frac{D_f - W}{ m}

=> a =  \frac{1085 - 968}{  98.8}

=> a = 1.1842 \ m/s^2

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