Question

An airplane is flying at a speed of 45 m/s when it drops a 40 kg food package to the Polar exploration team. If the plane drops the package from an altitude of 120 m and air resistance can be neglected, how fast is the package moving when it hits the ground?

Answer 1

To solve this problem we will apply the concepts related to energy conservation, so the potential energy in the package must be equivalent to its kinetic energy. From there we will find the speed of the package in the vertical component. The horizontal component is given, as it is the same as the one the plane is traveling to. Vectorially we will end up finding its magnitude. So,

$PE = KE$

$mgh = \frac{1}{2}mv^2$

Here,

m = Mass

g = Gravity

h = Height

v = Velocity

Rearranging to find the velocity

$v = \sqrt{2gh}$

Replacing,

$v = \sqrt{2(9.8)(120)}$

$v = 48.49m/s$

Using the vector properties the magnitude of the velocity vector would be given by,

$|V| = \sqrt{v_x^2+v_y^2}$

$|V| = \sqrt{45^2+48.42^2}$

$|V| = 66.2m/s$

Therefore the package is moving to 66.2m/s