Answer:
Approximately [tex]2.1\; \rm km[/tex], assuming that [tex]g = -9.8\; \rm m \cdot s^{-2}[/tex].
Explanation:
Let [tex]t[/tex] denote the time required for the package to reach the ground. Let [tex]h(\text{initial})[/tex] and [tex]h(\text{final})[/tex] denote the initial and final height of this package.
[tex]\displaystyle h(\text{final}) = \frac{1}{2}\, g\, t^2 + h(\text{initial})[/tex].
For this package:
Solve for [tex]t[/tex], the time required for the package to reach the ground after being released.
[tex]\displaystyle t^{2} = \frac{2\, (h(\text{final}) - h(\text{initial}))}{g}[/tex].
[tex]\begin{aligned} t &= \sqrt{\frac{2\, (h(\text{final}) - h(\text{initial}))}{g} \\ &\approx \sqrt{\frac{2\times (0\; \rm m - 2500\; \rm m)}{(-9.8\; \rm m \cdot s^{-2})}} \approx 22.588\; \rm s\end{aligned}[/tex].
Assume that the air resistance on this package is negligible. The horizontal ("forward") velocity of this package would be constant (supposedly at [tex]95\; \rm m \cdot s^{-1}[/tex].) From calculations above, the package would travel forward at that speed for about [tex]22.588\; \rm s[/tex]. That corresponds to approximately:[tex]95\; \rm m \cdot s^{-1} \times 22.588\; \rm s \approx 2.1 \times 10^{3}\; \rm m = 2.1\; \rm km[/tex].
Hence, the package would land approximately [tex]2.1\; \rm km[/tex] in front of where the plane released the package.
