(a) 240,000 N
We can solve the problem by using the impulse theorem, which states that the impulse exerted on the man is equal to its change in momentum:
[tex]I=\Delta p\\F\Delta t= m(v-u)[/tex]
where
F is the average net force exerted on the man
[tex]\Delta t=2.0 ms=0.002 s[/tex] is the contact time
m = 75 kg is the man's mass
v=0 m/s is the final speed of the man
u = 6.4 m/s is his initial speed
Solving the formula for F, we find
[tex]F=\frac{m(v-u)}{\Delta t}=\frac{(75 kg)(0-(6.4 m/s)}{0.002 s}=-240,000 N[/tex]
And the negative sign simply means that the force (upward) is opposite to the initial velocity of the man (downward).
(b) 4800 N
The problem can be solved exactly as the previous part, but this time the time of contact is
[tex]\Delta t=0.10 s[/tex]
So the average force is
[tex]F=\frac{m(v-u)}{\Delta t}=\frac{(75 kg)(0-(6.4 m/s)}{0.10 s}=-4800 N[/tex]
(c) 239,265 N and 4065 N
The force found in part (a) and (b) corresponds to the net force, which is the difference between the force exerted by the ground on the man (N, upward), and the weight of the man (W, downward):
[tex]F=N-W[/tex]
For part (a), we can find N as
[tex]N=F+W=240,000 N-(mg)=240,000 N-(75 kg)(9.8 m/s^2)=239,265 N[/tex]
While for part (b), we find
[tex]N=F+W=4800 N-(mg)=4800 N-(75 kg)(9.8 m/s^2)=4065 N[/tex]
