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

Approximately
`2.1\; \rm km`, assuming that
`g = -9.8\; \rm m \cdot s^(-2)`.

Step-by-step 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}) = (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{(2* (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) * 22.588\; \rm s \approx 2.1 * 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.