Chapter 3

Q. 3.11

CASE STUDY Clearing the Trees in the Brakes Configuration

When Reese saw that she was in danger of hitting the trees, she set her parachute in the brakes configuration (case study, page 60). In this configuration, her velocity was constant at \overrightarrow{ v }=(6.0 \hat{\imath}-1.5 \hat{\jmath})  m / s Had she deployed her parachute at 750 m, would she have cleared the trees?



Verified Solution

Our plan is to use Reese’s velocity to find her horizontal displacement during the descent. If her horizontal displacement is at least 3500 m  (\Delta x \geq 3500  m ), she would safely clear the trees.

Because Reese’s velocity is constant, her average velocity and her instantaneous velocity are equal.

\begin{aligned}\vec{v}_{ av } & =\overrightarrow{ v }=(6.0 \hat{ \imath }-1.5 \hat{ \jmath })  m / s \\ v_{ av , x} & =6.0  m / s \\v_{ av , y} & =-1.5  m / s\end{aligned}

If she deploys her parachute at 750 m, then \Delta y = – 750  m for the jump. Her horizontal displacement \Delta x depends on time \Delta t it takes her to descend straight down. Find \Delta t from the known v_{av, y} and ∆y using Equation 3.25.

\begin{aligned}v_{ av , y} & =\frac{\Delta y}{\Delta t} \quad \quad (3.25) \\ \Delta t & =\frac{\Delta y}{v_{ av , y}}=\frac{-750  m }{-1.5  m / s }=500  s \end{aligned}

Use this \Delta t to find \Delta x. Equation 3.25 works in this case because the x component of velocity is also constant.

\begin{aligned}v_{ av , x} & =\frac{\Delta x}{\Delta t}\quad \quad (3.25) \\ \Delta x & =v_{ av , x} \Delta t=(6.0  m / s )(500  s ) \\\Delta x & =3000  m\end{aligned}

No, she would have hit the trees.

We conclude that Reese would have landed in the trees had she waited and opened her parachute at 750 m. The advantage of deploying her parachute at a higher altitude is that the longer she spends in the air, the larger the value of \Delta x, assuming that she has the parachute open in the brakes configuration.