Question

A rescue pilot drops a survival kit while her plane is flying at an ultitude of 2500m with a forward velocity of 95m. If the air friction is disregarded, how far advance of the starving explorer’s drop zone should she release the package?

Answer 1

Answer:

Approximately $2.1\; \rm km$, assuming that $g = -9.8\; \rm m \cdot s^{-2}$.

Explanation:

Let $t$ denote the time required for the package to reach the ground. Let $h(\text{initial})$ and $h(\text{final})$ denote the initial and final height of this package.

$\displaystyle h(\text{final}) = \frac{1}{2}\, g\, t^2 + h(\text{initial})$.

For this package:

• Initial height: $h(\text{initial}) = 2500\; \rm m$.
• Final height: $h(\text{final}) = 0\; \rm m$ (the package would be on the ground.)

Solve for $t$, the time required for the package to reach the ground after being released.

$\displaystyle t^{2} = \frac{2\, (h(\text{final}) - h(\text{initial}))}{g}$.

$\begin{aligned} t &= \sqrt{\frac{2\, (h(\text{final}) - h(\text{initial}))}{g} \\ &\approx \sqrt{\frac{2\times (0\; \rm m - 2500\; \rm m)}{(-9.8\; \rm m \cdot s^{-2})}} \approx 22.588\; \rm s\end{aligned}$.

Assume that the air resistance on this package is negligible. The horizontal ("forward") velocity of this package would be constant (supposedly at $95\; \rm m \cdot s^{-1}$.) From calculations above, the package would travel forward at that speed for about $22.588\; \rm s$. That corresponds to approximately:$95\; \rm m \cdot s^{-1} \times 22.588\; \rm s \approx 2.1 \times 10^{3}\; \rm m = 2.1\; \rm km$.

Hence, the package would land approximately $2.1\; \rm km$ in front of where the plane released the package.