Answer:
The package should be dropped at a distance of 758.4 m from the scientist.
Explanation:
As shown in the given figure, let us assume that when the plane is just above the point 'O' it drops the package and 'A' be the position of the scientist, where [tex]OA = x[/tex].
Given, the velocity of the plane (the x-component) [tex]V_{x} = 160m s^{-1}[/tex], at point 'O' [tex]V_{y} = 0[/tex], the vertical height of the plane at 'O' is [tex]y = 110~m[/tex].
If '[tex]t[/tex]' be the time required for dropping the package to the scientist at point 'A', and 'g' be the acceleration due to gravity then we can write,
[tex]&& -110 = v_{y}t - \dfrac{1}{2}gt^{2}\\&or,& -100 = - \dfrac{1}{2} \times 9.8 t^{2}\\&or,& t = \sqrt{\dfrac{2 \times 110~m}{9.8~m~s^{-2}}} \approx 4.74~s[/tex]
If 'x' be the distance from the scientist where the package from the plane should be dropped then
[tex]x = V_{x} \times t = 160~m~s^{-1} \times 4.74~s = 758.4~m[/tex]
