Question 3.8: CASE STUDY Magnitude and Direction to Miss the Trees In Exam......
CASE STUDY Magnitude and Direction to Miss the Trees
In Example 3.2, we drew a representation of Reese’s displacement from the time she set her parachute in the brakes configuration to the moment she landed safely in the field. In this example, we will find the magnitude and direction of her displacement assuming she just barely missed the trees as shown in Figure 3.1. In the next section, we will determine whether she would have survived the jump had she opened her parachute at the usual 750 m instead of at 1200 m.
A. Find the magnitude and direction of Reese’s displacement if she just barely misses the trees. Use the distance data given in Figure 3.1, page 60.
B. Find the magnitude and direction of Reese’s displacement if she opens her parachute at an altitude of 750 m instead of 1200 m. Assume she still covers the same horizontal distance as in part A.
A.
INTERPRET and ANTICIPATE
Roughly sketch Reese’s displacement \Delta \vec{r} in a two-dimensional coordinate system and show the vector components \Delta \vec{x} and \Delta \vec{y} of \Delta \vec{r}. Also show the angles \theta and \alpha, either of which describes the direction of the displacement vector (Fig. 3.36).
SOLVE
We are given \Delta x and \Delta y. Assign \Delta y a negative value (downward).
Modify A=\sqrt{A_x^2+A_y^2} (Eq. 3.12) to find the magnitude of the displacement.
\begin{aligned}& \Delta r=\sqrt{\Delta x^2+\Delta y^2}\quad \quad (3.12) \\& \Delta r=\sqrt{(3500 m )^2+(-1200 m )^2} \\& \Delta r=3.7 \times 10^3 m\end{aligned}Modify \theta=\tan ^{-1} A_y / A_x (Eq. 3.14) to find the direction.
\tan ^{-1} \frac{\Delta y}{\Delta x}=\tan ^{-1} \frac{-1200}{3500}=-19^{\circ}From our drawing, we see that we have just found \alpha. . To find \theta , we need to add 360° to our \alpha value.
\begin{aligned}& \alpha=-19^{\circ} \\& \theta=360^{\circ}+\alpha=360^{\circ}-19^{\circ} \\& \theta=341^{\circ}\end{aligned}CHECK and THINK
The answer expressed as \alpha = −19° means that Reese fell through an angle of 19° from the horizontal after she set her parachute in the brakes configuration at an altitude of 1200 m.
B.
INTERPRET and ANTICIPATE
With her parachute closed, Reese falls straight down so that when she opens her parachute the point directly below her is still 3500 m from the edge of the clearing. Her altitude is only 750 m, however.
SOLVE
This situation is like part A except that the y component of displacement has changed.
As before, modify Equation 3.12 to find the magnitude of the displacement.
\begin{aligned}& \Delta r=\sqrt{\Delta x^2+\Delta y^2}=\sqrt{(3500 m )^2+(-750 m )^2} \\& \Delta r=3.6 \times 10^3 m\end{aligned}Modify Equation 3.14 to find the direction in terms of \alpha.
\alpha=\tan ^{-1} \frac{\Delta y}{\Delta x}=\tan ^{-1}\left(\frac{-750} {3500}\right)=-12^{\circ}To find \theta we need to add 360° as in part A.
\begin{aligned}& \theta=360^{\circ}+\alpha=360^{\circ}-12^{\circ} \\ & \theta=348^{\circ}\end{aligned}CHECK and THINK
When we return to this CASE STUDY in the next section, we will determine whether or not Reese would have cleared the trees with her chute opening at 750 m.
