Respuesta :
Given Data:
*The speed of the plane is:
[tex]v=50\text{ m/s}[/tex]*The height from which the package is dropped is:
[tex]h=160\text{ m}[/tex]*The time at which the second package is dropped is:
[tex]t_2=t_1+2\text{ s}[/tex]Here,
[tex]t_1[/tex]is the time at which the first package is dropped.
Explanation:
The initial vertical velocity is:
[tex]v_y=0[/tex]The acceleration of the motion in vertical y-axis is:
[tex]a=g=9.8\text{ m/s}^2[/tex]Using the kinematics equation, we get:
[tex]h=v_yt_1+\frac{1}{2}a(t_1)^{2_}[/tex]Substituting the known values, we get:
[tex]\begin{gathered} h=0+\frac{1}{2}g(t_1)^2 \\ h=\frac{1}{2}g(t_1)^2 \\ t_1=\sqrt{\frac{2h}{g}} \\ =\sqrt{\frac{2(160\text{ m})}{9.8\text{ m/s}^2}} \\ =5.71\text{ s} \end{gathered}[/tex]The displacement of the first package is:
[tex]\begin{gathered} d_1=v\cdot t_1 \\ =50\text{ m/s}\times5.71\text{ s} \\ =285.7\text{ m} \end{gathered}[/tex]The time taken by the second package is:
[tex]\begin{gathered} t_2=2s\text{ + 5.71 s} \\ =7.71\text{ s} \end{gathered}[/tex]The displacement of the second package is:
[tex]\begin{gathered} d_2=v\cdot t_2 \\ =50\text{ m/s}\times7.71\text{ s} \\ =385.5\text{ m} \end{gathered}[/tex]The distance between the two packages when they are both dropped on the ground is:
[tex]\begin{gathered} D=d_2-d_1 \\ =385.5\text{ m - 285.7 m} \\ \approx100\text{ m} \end{gathered}[/tex]Final Answer:
The correct option is 100 m.
