A 70.0 -kg stuntman free falls from a building for 125 m and hits an airbag. The airbag exerts a force on him as he depresses it 5.0 m and comes to a complete stop. What is the average force exerted by the airbag? A. 1.0 x 102 N B. 2.8 x 102 N C. 1.7 x 104 N D. 2.8 x 104 N E. 1.0 x 104 N Group of answer choices

E. 1 x 10⁴ N

Explanation:

First we will calculate the final velocity of man when he hits the air bag. For that we apply third equation of motion to the motion from the top of building to air bags:

2as = Vf² - Vi²

where,

a = acceleration due to gravity = g = 9.8 m/s²

s = height = h 125 m

Vf = Final Speed = ?

Vi = Initial Speed = 0 m/s

Therefore,

2(9.8 m/s²)(125 m) = Vf² - (0 m/s)²

Vf = √(2450 m²/s²)

Vf = 49.5 m/s

Now, we apply the 3rd equation of motion to the decelerating motion of air bag:

2as = Vf² - Vi²

where,

a = ?

s = 5 m

Vf = 0 m/s

Vi = 49.5 m/s

Therefore,

2(a)(5 m) = (0 m/s)² - (49.5 m/s)²

a = (-2450 m²/s²)/(10 m)

a = -245 m/s²

negative sign for deceleration

a = 245 m/s²

Using Newton's Second Law of Motion:

F = ma

where,

F = Average force Applied by Air bags

m = mass of man = 70 kg

a = deceleration = 245 m/s²

Therefore,

F = (70 kg)(245 m/s²)

F = 1.7 x 10⁴ N

So, the correct option is:

E. 1 x 10⁴ N

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