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 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

Answer 1

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

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

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

=> $a = 1.1842 \ m/s^2$