Question

A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 200 m above the glacier at a speed of 130 m/s .

How far short of the target should it drop the package?

Answer 1

First remember that tratta of a parabolic move.

The reference system is considered on the ground.

vertical position equation is given by:

$y=y_{0}\dotplus V_{0}\cdot t\dotplus \frac{1}{2}a_{y}\cdot t^{2}$

where;
$Y_{0}=200m$ Initial position, the position that.
$a_{y}=-g=-9.8\frac{m}{s^{2}}$ acceleration of gravity, negative because it points down.

$t$  time

$V_{0y}=0\frac{m}{s}$  Vertical component of the velocity is zero because the packet is dropped.

First we calculate the time it takes to reach the ground. To evaluate this equation in $Y=0m$

$0m=200m \dotplus 0\frac{m}{s}\cdot t\dotplus \frac{1}{2}\left (-9.8\frac{m}{s^{2}} \right )\cdot t^{2}$

Simplifying.

$0=200m -4.9 \frac{m}{s^{2}}\cdot t^{2}$

We find the expression for t.

$0=200m -4.9 \frac{m}{s^{2}}\cdot t^{2}\Rightarrow t= \sqrt{\frac{-200m}{-4.9\frac{m}{s^{2}}}}=6.39s$

And horizontally it behaves as a uniform rectilinear motion. We can use the following expression.

$V_{x}=\frac{X}{t}$

Where:
$V_{x}=130\frac{m}{s}$   Horizontal component of velocity which is constant.
$x$     horizontal distance walking while it is falling.
$t=6.39s$ time.

Finding the expression for the distance the package

$V_{x}=\frac{X}{t}\Rightarrow X=V_{x}\cdot t$

Evaluated numerically.

$X=V_{x}\cdot t=130\frac{m}{s}\cdot 6.39s=830.54m$ Horizontal distance that should launch the package for arrives just where are the scientists.