Chapter 6, Problem 39PQ

(a)

To determine

The maximum and minimum terminal speed for the skydivers.

Answer to Problem 39PQ

The maximum and minimum terminal speed of the skydivers are given below.

skydiver	Weight (N)	Length (m)	Radius (m)	Maximum (m/s)	Minimum (m/s)
A	$670$	$2.00$	$0.25$	$103$	$32.2$
B	$675$	$1.75$	$0.33$	$78.2$	$30.1$
C	$907$	$1.95$	$0.43$	$70.0$	$29.0$

Explanation of Solution

Write the formula for the terminal velocity of an object.

    $v_t=\sqrt{\frac{2mg}{C\rho A}}$ (I)

Here, $v_t$ is the terminal velocity, $m$ is the mass, $A$ is the cross-sectional area, $C$ is the drag coefficient, $\rho$ is the density of the medium.

When the skydiver is pointing head down, the terminal velocity is maximum. The $C$ value when the head is pointing down is $0.5$. The cross-sectional area will be that of a circle.

Re-write the equation (I) for maximum terminal speed.

    $\begin{array}{c}{v}_{t,\max }=\sqrt{\frac{2mg}{0.5\rho \pi r^2}}\\ =\sqrt{\frac{4W}{\rho \pi r^2}}\end{array}$ (II)

Here, $r$ is the radius of the cylinder, $v_{t,\max }$ is the maximum terminal velocity, $W$ is the weight of the skydiver.

When the skydiver is pointing belly to earth, the terminal velocity is minimum. The $C$ value when the belly is pointing down is $1$. The cross-sectional area is that of a rectangle, whose one side will be the diameter and another side will be equal to the length of the cylinder which model the body.

Re-write the equation (I) for minimum terminal speed.

    $\begin{array}{c}{v}_{t,\min }=\sqrt{\frac{2mg}{\rho \left(2rl\right)}}\\ =\sqrt{\frac{W}{\rho rl}}\end{array}$ (III)

Here, $l$ is the length of the cylinder, $v_{t,\min }$ is the minimum terminal velocity.

Conclusion:

Substitute $670\text{N}$ for $W$, $1.29{\text{kg/m}}^{\text{3}}$ for $\rho$, $0.25\text{m}$ for $r$ in equation (II) to determine the maximum terminal velocity of the diver A.

    $\begin{array}{c}{v}_{t,\max }=\sqrt{\frac{4\left(670\text{N}\right)}{\left(1.29{\text{kg/m}}^{\text{3}}\right)\pi {\left(0.25\text{m}\right)}^2}}\\ =103\text{m/s}\end{array}$

Substitute $670\text{N}$ for $W$, $1.29{\text{kg/m}}^{\text{3}}$ for $\rho$, $0.25\text{m}$ for $r$, $2.0\text{m}$ for $l$ in equation (II) to determine the maximum terminal velocity of the diver A.

    $\begin{array}{c}{v}_{t,\min }=\sqrt{\frac{670\text{N}}{\left(1.29{\text{kg/m}}^{\text{3}}\right)\left(0.25\text{m}\right)\left(2.0\text{m}\right)}}\\ =32.2\text{m/s}\end{array}$

Similarly the maximum and minimum terminal speed of the skydiver B and C can be calculated.

The table below shows the maximum and minimum terminal velocity of the all three skydivers.

skydiver	Weight (N)	Length (m)	Radius (m)	Maximum (m/s)	Minimum (m/s)
A	$670$	$2.00$	$0.25$	$103$	$32.2$
B	$675$	$1.75$	$0.33$	$78.2$	$30.1$
C	$907$	$1.95$	$0.43$	$70.0$	$29.0$

(b)

To determine

The order in which the skydivers should leave the plane in order to form the formation.

Answer to Problem 39PQ

The order that should be followed to make the formation is C, B, A.

Explanation of Solution

To form the formation of the figure 6.1 all the divers at some point should reach together. Since the diver C has the lowest terminal speed, he should leave the plane first followed by diver B and last diver A.

In the order C, B and A, the divers B and A can initially travel head down to reach to the next divers. Then they can face the belly to down.

Conclusion:

The order that should be followed to make the formation is C, B, A.

(c)

To determine

The time that the first skydiver has to wait after jumping to join the formation. The change in waiting time for the first skydiver to join formation if it takes $10\text{s}$ to achieve terminal speed.

Answer to Problem 39PQ

The time that the first skydiver has to wait after jumping to join the formation is $84\text{s}$. If it takes if $10\text{s}$ to achieve terminal speed, the wait time will be higher to achieve the terminal speed and to catch up the previous skydivers.

Explanation of Solution

The total wait time for the first skydiver is the total time required for the formation. It is equal to the time taken for the jump of skydiver B and A, then the time taken by the skydiver A to catch up with the skydiver C or the wait time for diver A.

The time gap between the divers to leave the plane is $30\text{s}$. Thus the total time taken for the jumping of skydivers B and A is $60\text{s}$.

Write the formula for the distance travelled by skydiver A.

    $d_A=v_{C,\min }\left(60\text{s+t}\right)$

Here, $d_A$ is the distance travelled by skydiver A, $v_{C,\min }$ is the minimum terminal speed of diver C, $t$ is the wait time of A.

Write the formula for the distance travelled by the skydiver C.

    $d_C=v_{A,\max }t$

Here, $v_{A,\max }$ is the maximum terminal speed of A, $d_C$ is the distance travelled by diver C.

Since both the divers travel the same distance.

    $v_{A,\max }t=v_{C,\min }\left(60\text{s+t}\right)$

Re-write the above equation to get an equation for $t$.

    $\begin{array}{c}{v}_{A,\max }t-v_{C,\min }t=v_{C,\min }\left(60\text{s}\right)\\ t=\frac{v_{C,\min }\left(60\text{s}\right)}{v_{A,\max }-v_{C,\min }}\end{array}$

Conclusion:

Substitute $29\text{m/s}$ for $v_{C,\min }$, $103\text{m/s}$ for $v_{A,\max }$.

  $\begin{array}{c}t=\frac{\left(29\text{m/s}\right)\left(60\text{s}\right)}{103\text{m/s}-29\text{m/s}}\\ =24\text{s}\end{array}$

Thus the total wait time for first skydiver is $60\text{s}+24\text{s}=84\text{s}$.

If it takes if $10\text{s}$ to achieve terminal speed, the wait time will be higher to achieve the terminal speed and to catch up the previous skydivers.