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3 years ago

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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

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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