(a) [tex]1.5\cdot 10^5 N[/tex]
According to the impulse theorem, the impulse exerted on the man during the impact is equal to his change in momentum:
[tex]I=\Delta p\\F \Delta t = m \Delta v[/tex]
where we have
F = magnitude of the average force
[tex]\Delta t=4.10 ms = 0.0041 s[/tex] is the contact time
[tex]m=72.7 kg[/tex] is the mass of the man
[tex]\Delta v = v_f -v_i = 0-(8.46 m/s)=-8.46 m/s[/tex] is the change in velocity of the man
Solving the formula for F, we find
[tex]F=\frac{m \Delta v}{\Delta t}=\frac{(72.7 kg)(-8.46 m/s)}{0.0041 s}=-1.5\cdot 10^5 N[/tex]
And the negative sign simply means the direction of the force is opposite to the initial velocity of the man (so, the force points upward).
(b) 3705 N
This part of the exercise is exactly identical to part (a), but here the contact time is much longer:
[tex]\Delta t=0.166 s[/tex]
Substituting into the equation, we find
[tex]F=\frac{m \Delta v}{\Delta t}=\frac{(72.7 kg)(-8.46 m/s)}{0.166 s}=-3705 N[/tex]
(c) 2992.5 N
We have two forces acting on the man:
- The force that the ground exerts on the man, pointing upward:
[tex]N=3705 N[/tex]
- The force of gravity (weight of the man), pointing downward:
[tex]W=mg=(72.7 kg)(9.8 m/s^2)=712.5 N[/tex]
Since the directions are opposite, the resultant force will be
[tex]F=N-W=3705 N-712.5 N=2992.5 N[/tex]
and the direction is upward.
