Chapter 3, Problem 65GP

To determine

To Calculate: Angle (to the horizontal) at which the car should be in sights when the packet is released.

Answer to Problem 65GP

The car should be in sights at an angle of $\underset{\_}{50°}$ .

Explanation of Solution

Given:

Velocity of the helicopter $=215\text{ km/hr}$

Velocity of car $=155\text{ km/hr}$

Vertical distance between helicopter and car $=78\text{ m}$

Formula used:

The following kinematic equations:

  $y=y_0+v_{y0}t+\frac{1}{2}{a}_y{t}^2$

Where, $y$ is displacement, $y_0$ is the initial position, $v_{y0}$ is the initial velocity in the vertical direction, t is time taken and $a_y$ ( $9.8{\text{ m/s}}^{\text{2}}$ ) is acceleration.

  $\Delta x=V_{x}t$

Where, $\Delta x$ is the horizontal distance, $V_x$ is velocity in the horizontal direction and t is time taken.

Horizontal angle is given by,

  $\theta ={\tan }^{-1}\left(\frac{\Delta y}{\Delta x}\right)$

Where, $\Delta y$ is vertical distance and $\Delta x$ is the horizontal distance.

Calculation:

The altitude of the helicopter is taken as origin. Displacement of the package, $y=78\text{ m}$ . Initial velocity is 0.

Time of flight for the package is calculated as:

  $\begin{array}{l}y=y_0+v_y{}_{0}t+\frac{1}{2}{a}_y{t}^2\\ 78=0+0t+\frac{1}{2}\left(9.8\right)t^2\\ 78=4.9t^2\\ t^2=\frac{78}{4.9}\\ t=\sqrt{\frac{78}{4.9}}\\ t=3.99\text{ s}\end{array}$

Therefore, time of flight for the package is 3.99 seconds.

The difference in horizontal speed of the helicopter and the car is calculated as:

  $\begin{array}{c}215\text{ km/hr}-155\text{ km/hr }=60\text{ km/hr}\\ =60\times \frac{5}{18}\\ =16.67\text{ m/s}\end{array}$

This gives the frame of reference in which the car is at rest. Therefore, the helicopter moves with a relative speed of 16.67 m/s.

Horizontal distance travelled by the package, relative to the stationary car is calculated as:

  $\begin{array}{l}\Delta x=V_{x}t\\ \Delta x=\left(16.67\right)\left(3.99\right)\\ \Delta x=66.5\text{ m}\end{array}$

Therefore, the angle at which the package must be released is calculated as:

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{\Delta y}{\Delta x}\right)\\ \theta ={\tan }^{-1}\left(\frac{78}{66.5}\right)\\ \theta =49.55°\\ \theta \approx 50°\end{array}$

Conclusion:

The car should be in sights at an angle of $\underset{\_}{50°}$ .