When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. A 75 kg man just before contact with the ground has a speed of 4.6 m/s.

(a) In a stiff-legged landing he comes to a halt in 2.1 ms. Find the average net force that acts on him during this time.
N
(b) When he bends his knees, he comes to a halt in 0.13 s. Find the average force now.
N
(c) During the landing, the force of the ground on the man points upward, while the force due to gravity points downward. The average net force acting on the man includes both of these forces. Taking into account the directions of these forces, find the force of the ground on the man in parts (a) and (b).
stiff legged landing
N
bent legged landing
N

Doesnt implulse= change of momentum?

Force*time= mass*change velocity

You are looking for force, with two different times. As far as I can determine, it is a straightforth calculation.

To determine the average net force acting on the man during his landing, we can apply Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

(a) In the case of a stiff-legged landing, the man comes to a halt in 2.1 ms (milliseconds), which is equivalent to 2.1 × 10^-3 seconds. To find the average net force, we first need to calculate the acceleration experienced by the man. We can use the formula:

acceleration = change in velocity / time

The change in velocity is given by the final velocity minus the initial velocity:

change in velocity = 0 - 4.6 m/s = -4.6 m/s

Substituting the values into the formula:

acceleration = (-4.6 m/s) / (2.1 × 10^-3 s) = -2181.8 m/s²

Since the man is coming to a halt, the acceleration is negative. Now, we can calculate the average net force using Newton's second law:

average net force = mass × acceleration

average net force = 75 kg × (-2181.8 m/s²) = -163,635 N

Therefore, the average net force that acts on the man during a stiff-legged landing is -163,635 N (pointing upwards).

(b) Similarly, for the bent-legged landing, the man comes to a halt in 0.13 seconds. Using the same process as above, we find:

change in velocity = 0 - 4.6 m/s = -4.6 m/s

acceleration = (-4.6 m/s) / (0.13 s) = -35.4 m/s²

average net force = 75 kg × (-35.4 m/s²) = -2,655 N

Hence, the average net force that acts on the man during a bent-legged landing is -2,655 N (pointing upwards).

(c) The force of the ground on the man is equal in magnitude but opposite in direction to the force due to gravity on the man.

In a stiff-legged landing, the force of the ground on the man is 163,635 N (pointing downwards), which is equal in magnitude to the force due to gravity on the man.

In a bent-legged landing, the force of the ground on the man is 2,655 N (pointing downwards), which is again equal in magnitude to the force due to gravity on the man.