A 75 kg man steps out a window and falls (from rest) 1.0 m to the sidewalk. What is his speed just before his feet strike the pavement? (b) If the man falls with his knees and ankles locked, the only cushion for his fall is an approximately 0.50 cm give in the pads of his feet. Calculate the average force exerted on him by the ground in this situation. Painful?

The answers to the questions are;

(a) His speed just before his feet strike the pavement is 4.429 m/s.

(b) The average force exerted on him by the ground in breaking his fall is 145.15 KN.

Explanation:

(a) To solve the question, we first list out the known variables as follows

Mass of man = 75 kg

Initial height above the sidewalk of man = 1.0 m

Speed just before his feet strike the pavement = Required

Since the height of fall and the mass is given and we are required to find the final speed before he strikes the pavement then we make use of the equation of motion as follows

v² = u² + 2·g·s

Where

v = Final velocity before hitting the ground = Required

u = Initial velocity = 0 m/s

g = Acceleration due to gravity = 9.81 m/s²

s = Initial height above the sidewalk = 1.0 m

Therefore

v² = (0 m/s)² + 2×9.81 m/s²×1.0 m

= 19.62 m²/s²

v =  = 4.429 m/s

his speed just before his feet strike the pavement is 4.429 m/s.

(b) Here we note that the kintic energy gained by the man just before the fall is broken by the sidewalk is given by

Kinetic Energy, KE = 1/2·m·v²

= 1/2×75 kg × (4.429 m/s)² = 735.75 J

Therefore since energy = Force × distance

That is the force experienced by the man on breaking the fall is

Force × Distance of the force = Initial kinetic energy gained by the man

Here, distance of the force, d = 0.50 cm = ‪0.005 m

Therefore (Force F) × (Distance d) = Initial kinetic energy gained by the man  

F×d = 735.75 J

F = (735.75 J)/d = (735.75 J)/(‪0.005 m) = 147,150.00 N

Therefore the average force exerted on him by the ground in this situation is 147,150.00 N or 145.15 KN.