Question

Spymaster Chris, flying a constant 208 km/h horizontally in a low-flying helicopter, wants to drop secret documents into her contact’s open car which is traveling 156 km/h on a level highway 78.0 m below. At what angle (with the horizontal) should the car be in her sights when the packet is released (Fig. 3–55)?

Short answer

The obtained value of the angle made by the car is $\theta =53.6°$.

Step-by-step solution

Step 1. Calculate the time for the package to fall

The frame of reference in this problem is the car at rest at the ground level. In the case of vertical motion, the level of the helicopter is considered as the origin, and the downward direction is considered as positive.

Given data:

The speed of Chris is $v_1=208\mathrm{km}/\mathrm{h}$.

The speed of the car is $v_2=156\mathrm{km}/\mathrm{h}$.

The height at which the helicopter is flying is $h=78\mathrm{m}$.

The relation from the equation of motion is given by:

$h=ut+\frac{1}{2}g{t}^2$

Here, u is the speed of the package whose value is zero, t is the time taken for the package to fall, and g is the gravitational acceleration.

On plugging the values in the above relation, you get:

$\begin{array}{c}78 \mathrm{m}=\left(0\right)t+\frac{1}{2}\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)t^2\\ t=3.98\mathrm{s}\end{array}$

Step 2. Calculate the horizontal distance moved by the package

The formula to calculate the distance can be written as:

$\begin{array}{l}y=v_{x}t\\ y=\left(v_1-v_2\right)t\end{array}$

On plugging the values in the above relation, you get:

$\begin{array}{l}y=\left(208\mathrm{km}/\mathrm{h}-156\mathrm{km}/\mathrm{h}\right)\left(\frac{1\mathrm{m}/\mathrm{s}}{3.6\mathrm{km}/\mathrm{h}}\right)\left(3.98\mathrm{s}\right)\\ y=57.4 \mathrm{m}\end{array}$

Step 3. Calculate the angle made by the car

The formula to calculate the angle can be written as:

$\tan \theta =\frac{h}{y}$

On plugging the values in the above relation, you get:

$\begin{array}{c}\tan \theta =\left(\frac{78\mathrm{m}}{57.4\mathrm{m}}\right)\\ \theta =53.6°\end{array}$

Thus, $\theta =53.6°$is the angle under the horizontal for the package.