To find the magnitude of the average force exerted on the person's torso by their legs during deceleration, we can use the following physics concepts:

Step 1: Calculate the deceleration

We know the initial velocity ($v_i = 8.4 \, \text{m/s}$), the final velocity ($v_f = 0 \, \text{m/s}$) as the torso comes to rest, and the distance over which deceleration occurs ($d = 0.66 \, \text{m}$).

Using the kinematic equation:

$v_f^2 = v_i^2 + 2ad$

Where:

• $v_f = 0 \, \text{m/s}$
• $v_i = 8.4 \, \text{m/s}$
• $d = 0.66 \, \text{m}$
• $a$ is the deceleration we need to find.

Rearrange to solve for $a$:

$a = \frac{v_f^2 - v_i^2}{2d}$

Substitute the known values:

$a = \frac{0^2 - (8.4)^2}{2 \times 0.66}$

$a = \frac{-70.56}{1.32} \approx -53.45 \, \text{m/s}^2$

Thus, the deceleration is approximately $a = 53.45 \, \text{m/s}^2$ (the negative sign indicates deceleration).

Step 2: Calculate the average force

We can now use Newton's second law to find the average force:

$F = ma$

Where:

• $m = 41 \, \text{kg}$ is the mass of the torso
• $a = 53.45 \, \text{m/s}^2$ is the deceleration

$F = 41 \times 53.45 \approx 2191.45 \, \text{N}$

Thus, the average force exerted on the torso by the legs during deceleration is approximately 2191.45 N.

Summary:

The magnitude of the average force exerted on the torso by the legs is 2191.45 N.

Would you like further details on the process?

Here are some related questions to consider:

1. What is the importance of deceleration distance in reducing injury?
2. How does mass affect the force experienced during deceleration?
3. Can you derive similar equations for different initial velocities?
4. How does bending the knees affect the overall deceleration experienced by the body?
5. What is the relationship between force and acceleration in this scenario?

Tip: The greater the distance over which deceleration occurs, the smaller the force experienced, which is why bending your knees helps absorb the impact when landing.