(a) The man comes to a halt in 1.99 ms = 1.99 × 10⁻³ s, so his average acceleration slowing him from 6.33 m/s to a rest is
<em>a</em> = (6.33 m/s - 0) / (1.99 × 10⁻³ s) ≈ 3180 m/s²
so that the average net force on the man during the landing is
<em>F</em> = (70.8 kg) <em>a</em> ≈ 225,000 N
i.e. with magnitude 225,000 N.
(b) With knees bent, the man has an average acceleration of
<em>a</em> = (6.33 m/s - 0) / (0.147 s) ≈ 43.1 m/s²
and hence an average net force of
<em>F</em> = (70.8 kg) <em>a</em> ≈ 3050 N
(c) The net force on the man is
∑ <em>F</em> = <em>n</em> - <em>w</em> = <em>m a</em>
where
<em>n</em> = magnitude of the normal force, i.e. the force of the ground pushing up on the man
<em>w</em> = the man's weight, <em>m g</em> ≈ 694 N
<em>m</em> = the man's mass, 70.8 kg
<em>g</em> = mag. of the acceleration due to gravity, 9.80 m/s²
<em>a</em> = the man's acceleration
Using the acceleration in part (b), we have
<em>n</em> = <em>m g</em> + <em>m a</em> = <em>m</em> (<em>g</em> + <em>a</em>)
<em>n</em> = (70.8 kg) (9.80 m/s² + 43.1 m/s²) ≈ 3740 N
