Answer:
Assume that the air resistance on the supplies is negligible, and that [tex]g = 9.8 \; {\rm m\cdot s^{-2}}[/tex].
The plane need to drop the supplies when it is horizontally approximately [tex]700\; {\rm m}[/tex] away from the hill.
The supplies will hit the tree.
Explanation:
Let [tex]u_{y}[/tex] and [tex]v_{y}[/tex] denote the initial and final vertical velocity of the supply; [tex]u_{y} = 0\; {\rm m\cdot s^{-1}}[/tex] since the plane was flying horizontally.
Let [tex]x_{y}[/tex] denote the vertical displacement of the supply; [tex]x_{y} = 350\; {\rm m} - 810\; {\rm m} = (-460)\; {\rm m}[/tex].
Let [tex]a_{y}[/tex] denote the vertical acceleration of the supply; [tex]a = (-g) = (-9.8)\; {\rm m\cdot s^{-2}}[/tex].
Make use of the SUVAT equation [tex]{v_{y}}^{2} - {u_{y}}^{2} = 2\, a_{y}\, x_{y}[/tex] to find [tex]v_{y}[/tex], the final vertical velocity of the supply:
[tex]\begin{aligned} {v_{y}}^{2} &= {u_{y}}^{2} + 2\, a_{y}\, x_{y} \end{aligned}[/tex].
[tex]\begin{aligned} v_{y} &= -\sqrt{{u_{y}}^{2} + 2\, a_{y}\, x_{y}} \\ &= -\sqrt{0^{2} + 2\, (-9.8)\, (-460)}\; {\rm m\cdot s^{-1}} \\ &\approx (-94.953)\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].
(Negative since the supply would be travelling downwards.)
Let [tex]t[/tex] denote time it takes for the supply to land on the hill after being dropped from the plane. Make use of the SUVAT equation [tex]t = (v_{y} - u_{y}) / (a)[/tex] to find the value of [tex]t\![/tex]:
[tex]\begin{aligned} t &= \frac{v_{y} - u_{y}}{a} \\ &\approx \frac{(-94.953) - 0}{(-9.8)} \; {\rm s}\\ &\approx 9.6890 \; {\rm s} \end{aligned}[/tex].
Apply unit conversion and ensure that [tex]v_{x}[/tex], the horizontal speed of the plane is in the standard unit [tex]{\rm m\cdot s^{-1}}[/tex]:
[tex]\begin{aligned} v_{x} &= \frac{260\; {\rm km}}{1\; {\rm h}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \\ &\approx 72.222\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Under the assumptions, the horizontal speed of the supply will be the same as that of the plane- [tex]v_{x} \approx 72.222\; {\rm m\cdot s^{-1}}[/tex]- until it lands.
While in the air, the supply will travel a horizontal distance of:
[tex]\begin{aligned}x_{x} &= v_{x}\, t \\ &\approx 72.222\; {\rm m\cdot s^{-1}} \times 9.6890\; {\rm s} \\ &\approx 699.76\; {\rm m}\end{aligned}[/tex].
Hence, for the supply to land exactly at the top of the hill, the plane need to drop the supply while at a horizontal distance of approximately [tex]700\; {\rm m}[/tex] away from the hill.
The horizontal distance between the trees and the location where the plane dropped the supply would be approximately [tex](700\; {\rm m} - 30\; {\rm m}) = 670\; {\rm m}[/tex]. The time required for the the supply to reach that horizontal position would be:
[tex]\begin{aligned} t &= \frac{x_{x}}{v_{x}} \approx \frac{669.76\; {\rm m}}{72.222\; {\rm m\cdot s^{-1}}} \approx 9.2736\; {\rm s}\end{aligned}[/tex].
Let [tex]h_{0}[/tex] denote the initial height of the supply (relative to the sea level.) In this question, [tex]h_{0} = 810\; {\rm m}[/tex].
Let [tex]h(t)[/tex] denote the height of the supply (relative to the sea level) after being dropped from the plane for time [tex]t[/tex].
The SUVAT equation [tex]h(t) = (1/2)\, a\, t^{2} + u_{y}\, t + h_{0}[/tex] gives an expression for [tex]h(t)[/tex]. Make use of this equation to find the height of the supply (relative to the sea level) when the supply reach the horizontal position of the trees at [tex]t \approx 9.2736\; {\rm s}[/tex]:
[tex]\begin{aligned} h(t) &= \frac{1}{2}\, a\, t^{2} + u_{y}\, t + h_{0} \\ &= \frac{1}{2}\times (-9.8)\, (9.2736)^{2} + 0\times 9.2736 + 810 \\ &\approx 388.60\; {\rm m} \end{aligned}[/tex].
Note that the altitude of the top of the trees is [tex]350\; {\rm m} + 40\; {\rm m} = 390\; {\rm m}[/tex] relative to the sea level. Since [tex]388.90\; {\rm m} < 390\; {\rm m}[/tex], the supplies will run into the trees.
