A hot air balloon is hovering in the air when it drops a 40 Kg food package to some lost golfers. If the package is dropped from an altitude of 500 m and air resistance can be neglected, how fast is the package moving when it hits the ground?

We can calculate this with the law of conservation of energy. Here we have a food package with a mass m=40 kg, that is in the height h=500 m and all of it's energy is potential. When it is dropped, it's potential energy gets converted into kinetic energy. So we can say that its kinetic and potential energy are equal, because we are neglecting air resistance:

Ek=Ep, where Ek = (1/2) * m*v² and Ep=m*g*h, where m is the mass of the body, g=9.81 m/s² and h is the height of the body.

(1/2) * m*v²=m*g*h, masses cancel out and we get:

(1/2) * v²=g*h, and we multiply by 2 both sides of the equation

v²=2*g*h, and we take the square root to get v:

v=√ (2*g*h)

v=99.04 m/s

So the package is moving with the speed of v = 99.04 m/s when it hits the ground.