A helicopter is flying in a straight line over a level field at a constant speed of $\mathbf{6}\mathbf{.}\mathbf{2}\mathbf{m}\mathbf{/}\mathbf{s}$and at a constant altitude of$\mathbf{9}\mathbf{.}\mathbf{50}\mathbf{m}$. A package is ejected horizontally from the helicopter with an initial velocity of$\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}\mathbf{s}$relative to the helicopter and in a direction opposite the helicopter’s motion. (a) Find the initial speed of the package relative to the ground. (b) What is the horizontal distance between the helicopter and the package at the instant the package strikes the ground? (c) What angle does the velocity vector of the package meet with the ground at the instant before impact, as seen from the ground?

• Speed of helicopter,${\mathrm{V}}_{\mathrm{ix}1}=6.20\mathrm{m}/\mathrm{s}$ .
• Altitude at which the helicopter is flying,$\mathrm{h}=9.50\mathrm{m}$.
• Initial velocity of the ejected package, ${\mathrm{V}}_{\mathrm{ix}2}=12.0\mathrm{m}/\mathrm{s}$.

The motion of the package is projectile motion; we can find the initial speed of the package by subtracting the velocities which are opposite.

Using a kinematic equation we can find time from vertical motion and using that time we can find the horizontal distance.

Horizontal velocity remains the same; we need to find the vertical velocity at the ground.

Using that velocity component we can find the angle.

Formulae:

The first kinematic equation of motion, $v_f=v_i+at$

Where is the final velocity, $v_i$ is the initial velocity, a is the acceleration, and t is the time.

The second kinematic equation of motion, $∆x=v_{0}t+\left(\frac{1}{2}\right)\times a\times t^2$

Where $∆x$is the distance.

The relative speed of the package w.r.t. ground can be given as:

$Vi=V_{ix2}-V_{ix1}$

Substitute the values in the above equation.

$$\begin{gathered} =12\mathrm{m}/\mathrm{s}-6.2\mathrm{m}/\mathrm{s} \\ =5.8\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the horizontal distance is 17m

For finding the horizontal distance, we need to find the time from vertical motion.

Initially vertical velocity is 0 because the package is thrown horizontally

Hence, using equation (ii), substitute the values and we get time as given:

$$\begin{gathered} -9.50\mathrm{m}/\mathrm{s}=0+\left(\frac{1}{2}\right)(-9.8\mathrm{m}/{\mathrm{s}}^2)t^2 \\ {t}^2=\frac{9.50}{4.9} \\ =1.9387{\mathrm{s}}^2 \\ t=1.39s \end{gathered}$$

Using this time, we can find horizontal distance as follows:

$$\begin{gathered} x=12(1.39) \\ =16.70\mathrm{m} \\ =17\mathrm{m} \end{gathered}$$

Hence, the value of the horizontal distance is$17\mathrm{m}$

The horizontal velocity of the package remains the same. Hence,

$$\begin{gathered} V_{fx}=V_{ix} \\ =5.8\mathrm{m}/\mathrm{s} \end{gathered}$$

Again, using equation (i), we get the vertical velocity at the ground as follows:

$$\begin{gathered} {\mathrm{V}}_{\mathrm{fy}}=0+(-9.8)(1.39) \\ =13.621\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the angle at which the velocity vector makes an impact before meeting the ground is given as

$$\begin{gathered} \tan \theta =\frac{V_{fy}}{V_{fx}} \\ \theta ={\tan }^{-1}\left(\frac{V_{fy}}{V_{fx}}\right) \\ ={\tan }^{-1}\left(\frac{13.621}{5.8}\right) \\ =66.93 \\ =67° \end{gathered}$$

Hence, the value of the angle is$67°$