Answer
a) $7.67\;\rm m/s$
b) $4500\;\rm N$
c) $1.9\times 10^5\;\rm N$
d) $7.5\times10^6\;\rm N/m^2$; the bone will not break.
e) $3.17\times10^8\;\rm N/m^2$; the bone will break.
Work Step by Step
a) In this part, the author asks to find the speed of the person just before landing by using chapter 2.
We know that $v_{iy}=0\;\rm m/s$, $y_i=3\;\rm m$, $y_f=0\;\rm m$, and $g=9.8\;\rm m/s^2$; whereas $i\rightarrow$ means initial and $f\rightarrow$ means final.
we can use the kinematic formula of
$$v_{fy}^2=v_{iy}^2+2a_y\left(y_f-y_i\right)$$
Plugging the known and solving for $v_f$;
$$v_{fy} =\pm\sqrt{0^2+\left[2\cdot (-9.8)\cdot\left(0-3\right)\right]}$$
Since the author asks about the speed, not the velocity, we can ignore the direction of this final speed and just find its magnitude.
$$\boxed{v_{fy} =\bf{7.67}\;\rm m/s}$$
b)
When the person's foot touches the ground there will be two forces acting on him. The first force is his own weight due to Earth's gravitational pull (downward) and the normal force from the ground (upward), see the figure below.
Thus,
$$\sum F_y=F_n-mg=ma_y\tag 1$$
whereas $a_y$ is the decelerating that brings him to stop.
We also know that the work done on the person by the net force during this stop process is given by
$$W=\sum F_y d$$
whereas $d=0.5$ m which is the distance traveled by his Center of Gravity downward.
Noting that the work done is equal to the change in the kinetic energy. So
$$\sum F_y d=-\Delta KE $$
$$\sum F_y d=-\left(\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2\right)$$
whereas $v_i$ is the person's speed just before his feet touch the ground (which we found in part (a) above).
Plugging the known and $\sum F_y$ from (1), and solving for $F_n$;
$$(F_n-mg)d=-\frac{1}{2}m\cdot0^2+\frac{1}{2}mv_i^2$$
$$(F_n-mg) d= \frac{1}{2}mv_i^2\tag 2$$
$$ F_n = 2\cdot\frac{1}{2}mv_i^2+mg$$
$$ F_n = [65\cdot 7.67^2]+[65\cdot 9.8]=4461\;\rm N$$
$$\boxed{F_n\approx \bf{4500}\;\rm N}$$
c) Applying the same process we did in part (b) and using equation (2) when $d=1\;\rm cm$.
Thus,
$$(F_n-mg) \cdot 0.01= \frac{1}{2}\cdot 65\cdot 7.67^2 $$
$$F_n=[50\cdot 65\cdot 7.67^2]+[65\cdot 9.8]=1.918\times 10^5\;\rm N$$
$$\boxed{F_n\approx \bf{1.9\times 10^5}\;\rm N}$$
d)
We know, from table 9-2, that the compressive strength of the tibia is $170\times 10^6\;\rm N/m^2$ and now we need to find the stress in the tibia in this situation.
$${\rm Stress\; in\; the\; tibia}=\dfrac{F}{A}$$
Plugging $F$ from part (b) and $A$ from the given:
$${\rm Stress\; in\; the\; tibia}=\dfrac{4500}{3\times10^{-4}}=150\times10^5\;\rm N/m^2$$
This is the total stress in both legs, and we need to find the stress in one leg (and hence one tibia);
Thus,
$${\rm Stress\; in\; one\; tibia}=\dfrac{150\times10^5}{2}=750\times10^4\;\rm N/m^2$$
which is less than the compressive strength of the tibia of $170\times 10^6\;\rm N/m^2$.
Thus the bone will not break.
e) Repeating the same approach, as we did in (d) but using the force we found in part (c);
$${\rm Stress\; in\; the\; tibia}=\dfrac{F}{A}$$
Plugging $F$ from part (c) and $A$ from the given:
$${\rm Stress\; in\; the\; tibia}=\dfrac{1.9\times 10^5}{3\times10^{-4}}=6.33\times10^9\;\rm N/m^2$$
This is the total stress in both legs, and we need to find the stress in one leg (and hence one tibia);
Thus,
$${\rm Stress\; in\; one\; tibia}=\dfrac{6.33\times10^9}{2}=3.17\times10^8\;\rm N/m^2$$
which is more than the compressive strength of the tibia of $170\times 10^6\;\rm N/m^2$.
Thus the bone will break.
