An apple weighing 200 g falls freely from an apple tree branch at a height of 2 m to the ground.

An apple weighing 200 g falls freely from an apple tree branch at a height of 2 m to the ground. What will be the kinetic energy of an apple before hitting the ground? What will be its speed before hitting?

To begin with, let’s determine what the potential energy of the apple will be at the moment when it still weighs on the tree:

En = m * g * h;

Ep = 0.2 * 10 * 2 = 4.

As we know from the school course, before hitting the ground, all potential energy will go into kinetic, that is:

Ek = 4.

Now we can determine what the speed of the apple will be equal to before it hits the ground:

Ek = m * v ^ 2/2;

v ^ 2 = 2 * Ek: m;

v = √ (2 * Ek: m);

v = √ (2 * 4: 0.2);

v = √40;

v = 2√10.

Answer: Ek = 4 J, v = 2√10 m / s.