Question

A pear hangs in a tree at a height of 1.8 m. The pear has a mass of 0.2 kg. The pear falls out of the tree and lands on the ground.
1. How much kinetic energy does the pear have when it reaches the ground?
2. How fast is the pear moving when it hits the ground?

Answer 1

We apply the simple principle of energy conservation. At height $h=1.8 m$ the pear has potential energy (and no kinetic energy). When it hits the ground the pear has kinetic energy (and no potential energy). Because the pear falls in a conservative force field (the gravitational field) the initial potential energy = final kinetic energy.
$mgh=mv^2/2$
Thus its kinetic energy is
$E_k =2gh =2*9.81*1.8 =35.316 (J)$
and its speed is
$v= \sqrt{2gh} = \sqrt{2*9.81*1.8}=5.94 (m/s)$