We apply the simple principle of energy conservation. At height [tex]h=1.8 m[/tex] 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.
[tex]mgh=mv^2/2 [/tex]
Thus its kinetic energy is
[tex]E_k =2gh =2*9.81*1.8 =35.316 (J)
[/tex]
and its speed is
[tex]v= \sqrt{2gh} = \sqrt{2*9.81*1.8}=5.94 (m/s)
[/tex]
