91Ó°ÊÓ

• 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, $∆x=v_{0}t+\left(\frac{1}{2}\right)×a×t^2$

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

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

Hence, the value of the angle is$67°$