Question

A rescue pilot drops a survival kit while her plane is flying at a height of 1,635 m with a horizontal velocity of 193 m/s. If air friction is ignored, how far in advance of the starving explorer’s drop zone should she release the package? (find horizontal displacement [range] of the package)

Answer 1

Let's write the equations of motion of a survival kit (or the package) in horizontal and vertical directions:

$$x=v_{0x}t, (1)$$ $$y=v_{0y}t+\dfrac{1}{2}gt^2, (2)$$

here, $x$ is the horizontal displacement of the package, $v_{0x}=193\ \dfrac{m}{s}$ is the initial horizontal velocity of the package, $t$ is the time that the package takes to reach the ground, $y=1635\ m$ is the vertical displacement of the package, $v_{0y}=0\ \dfrac{m}{s}$ is the initial vertical velocity of the package, $g=9.8\ \dfrac{m}{s^2}$ is the acceleration due to gravity.

Let's first find the time that the package takes to reach the ground from the second equation:

$$t=\sqrt{\dfrac{2y}{g}}.$$

Finally, we can substitute $t$ into the first equation and calculate the horizontal displacement of the package:

$$x=v_{0x}\sqrt{\dfrac{2y}{g}}=193\ \dfrac{m}{s}\cdot\sqrt{\dfrac{2\cdot 1635\ m}{9.8\ \dfrac{m}{s^2}}}=3525\ m.$$

Answer:

$x=3525\ m.$