A 73.3 kg man steps out a window and falls (from rest) 1.09 m to a sidewalk. What is his speed just before his feet strike the pavement?

4.62 m/s

If the man falls with his knees and ankles locked, the only cushion for his fall is a 0.510 cm give in the pads of his feet. Calculate the average force exerted on him by the ground in this situation. This average force is sufficient to cause cartilage damage in the joints or to break bones.

4.62 m/s is correct for the impact speed.

For the average force Fav, use:

Fav*X = (1/2)M V^2
where V is the velocity at impact and X is the 5.1*10^-2 m compression distance required to stop.

Fav = 15,300 N

To find the man's speed just before his feet strike the pavement, we can use the equation of motion for an object in free fall:

v^2 = u^2 + 2as

where:
v = final velocity (unknown)
u = initial velocity (0 m/s, since the man falls from rest)
a = acceleration due to gravity (9.8 m/s^2)
s = distance fallen (1.09 m)

Plugging in the values:

v^2 = 0^2 + 2(9.8)(1.09)
v^2 = 0 + 21.368
v^2 = 21.368
v ≈ √21.368
v ≈ 4.62 m/s

So, the man's speed just before his feet strike the pavement is approximately 4.62 m/s.

Now, let's calculate the average force exerted on him by the ground when his knees and ankles are locked. We can use the impulse-momentum principle:

Favg * t = Δp

where:
Favg = average force (unknown)
Δp = change in momentum (mass * velocity)
t = time taken to come to a stop (unknown)

Since we know the man's mass (73.3 kg) and final velocity (0 m/s), we can calculate the initial momentum:

p = m * v
p = 73.3 * 4.62
p ≈ 338.5866 kg m/s

The time taken to come to a stop can be found using the equation of motion:

v = u + at

where:
v = final velocity (0 m/s)
u = initial velocity (4.62 m/s)
a = acceleration (which is caused by the average force)
t = time taken to come to a stop (unknown)

Rearranging the equation:

t = (v - u) / a

Substituting the values:

t = (0 - 4.62) / a

Now, the acceleration can be found using Newton's second law:

F = m * a

Rearranging the equation:

a = F / m

Substituting the given value for the give in the pads of his feet (0.510 cm = 0.0051 m), and the mass of the man (73.3 kg):

a = F / 73.3

Now, we can substitute this value of acceleration in the equation for time:

t = (0 - 4.62) / (F / 73.3)
t = -4.62 * (73.3 / F)
t = -338.5866 / F

Substituting back into the impulse-momentum equation:

Favg * (-338.5866 / F) = 338.5866

Simplifying:

Favg = -338.5866^2 / 338.5866
Favg ≈ -338.5866 N

The negative sign indicates that the force is in the opposite direction of the motion (upwards). However, the magnitude of the force is what we are interested in, so we can take the absolute value:

|Favg| ≈ 338.5866 N

Therefore, the average force exerted on the man by the ground in this situation is approximately 338.5866 Newtons.