Question

A plane flying horizontally at a speed of 50.0 m/s and at an elevation of 160 m drops a package. Two seconds later it drops a second package. How far apart will the two packages land on the ground? 100 m 162 m 177 m 283 m

Answer 1

Given Data:

*The speed of the plane is:

`v=50\text{ m/s}`

*The height from which the package is dropped is:

`h=160\text{ m}`

*The time at which the second package is dropped is:

`t_2=t_1+2\text{ s}`

Here,

`t_1`

is the time at which the first package is dropped.

Step-by-step explanation:

The initial vertical velocity is:

`v_y=0`

The acceleration of the motion in vertical y-axis is:

`a=g=9.8\text{ m/s}^2`

Using the kinematics equation, we get:

`h=v_yt_1+(1)/(2)a(t_1)^(2_)`

Substituting the known values, we get:

`\begin{gathered} h=0+(1)/(2)g(t_1)^2 \\ h=(1)/(2)g(t_1)^2 \\ t_1=\sqrt{(2h)/(g)} \\ =\sqrt{\frac{2(160\text{ m})}{9.8\text{ m/s}^2}} \\ =5.71\text{ s} \end{gathered}`

The displacement of the first package is:

`\begin{gathered} d_1=v\cdot t_1 \\ =50\text{ m/s}*5.71\text{ s} \\ =285.7\text{ m} \end{gathered}`

The time taken by the second package is:

`\begin{gathered} t_2=2s\text{ + 5.71 s} \\ =7.71\text{ s} \end{gathered}`

The displacement of the second package is:

`\begin{gathered} d_2=v\cdot t_2 \\ =50\text{ m/s}*7.71\text{ s} \\ =385.5\text{ m} \end{gathered}`

The distance between the two packages when they are both dropped on the ground is:

`\begin{gathered} D=d_2-d_1 \\ =385.5\text{ m - 285.7 m} \\ \approx100\text{ m} \end{gathered}`

Final Answer:

The correct option is 100 m.