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Mike has a mass of 97 kg. He jumps out of a perfectly good airplane that is 2000 m above the ground. After he falls 1000 m, when his downward speed is 68 m/s, Mike opens his parachute. The positive y-direction is downward.

(a) Calculate the average magnitude of the upward force of the air resistance on Mike during his initial descent.

(b) After Mike opens his parachute, he continues to descend, eventually reaching the ground with a speed of 4.0 m/s. Calculate the average upward force during this part of Mike's descent.

(c) At the same time Mike jumps out of the airplane, his wallet (mass of 0.3 kg) falls out of his pocket. Calculate the wallet's downward speed when it reaches the ground. For this calculation, assume that air resistance is negligible.

Respuesta :

Answer:

a)  fr = 224.3 N , b)   fr = 224.3 N , c)   v = 198.0  m/s

Explanation:

a) For this exercise let's start by calculating the acceleration in the fall

             v² = v₀² - 2 a (y-y₀)

When it jumps the initial vertical speed is zero

             a = -v² / 2 (y-y₀)

             a = -68 2/2 (1000-2000)

             a = 2,312 m / s²

Let's use the second net law to enter the average friction force

            fr = m a

            fr = 97 2,312

            fr = 224.3 N

b) let's look for acceleration

            v² = v₀² - 2 a y

            a = (v² –v₀²) / 2 (y-y₀)

            a = (4² - 68²) / 2 (0-1000)

            a = 2,304 m / s²

            fr = m a

            fr = 97 2,304

            fr = 223.5 N

c) the speed of the wallet is searched with kinematics

           v² = v₀² - 2 g (y-y₀)

           v = √ (0-2 9.8 (0-2000))

           v = 198.0  m/s

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