91Ó°ÊÓ

The weight of the stone = 52.9 N

The initial speed of the stone,v=20.0 m/s

The force of air drag,$F_f=0.265\mathrm{N}$

The stone going up the height gains potential energy and loses some part of the initial kinetic energy to the air drag. While coming down to the ground again, it converts the potential energy to kinetic energy and thermal energy. Thus, we need to applythelaw of conservation of energy in both parts of the motion.

Formula:

$$\begin{gathered} \mathrm{K}.\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2 \\ \mathrm{P}.\mathrm{E}=\mathrm{mgh} \\ {\mathrm{E}}_{\mathrm{th}}={\mathrm{F}}_{\mathrm{f}}\mathrm{d} \end{gathered}$$

Using the conservation of energy principle,

$$\begin{gathered} \left(KE\right)atbottom=\left(PE\right)ath+E_{th} \\ \frac{1}{2}m{v}^2=mgh+F_{f}h \\ \frac{1}{2}×\frac{5.29}{9.8}×{\left(20.0\right)}^2=\left(5.29+0.265\right)h \\ \left(5.55\right)h=\frac{5.29×400}{2×9.8} \\ h=19.4m \end{gathered}$$

Hence, the maximum height reached by the stone = 19.4 m

Now, when the stone starts coming down, the energy is converted to kinetic energy and thermal energy. So we apply the law of conservation of energy again as follows,

$$\begin{gathered} {\mathrm{PE}}_{\mathrm{top}}={\mathrm{E}}_{\mathrm{th}}+{\mathrm{KE}}_{\mathrm{f}} \\ {\mathrm{KE}}_{\mathrm{f}}={\mathrm{PE}}_{\mathrm{f}}-{\mathrm{E}}_{\mathrm{th}} \\ \frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2=\mathrm{mgh}-{\mathrm{F}}_{\mathrm{f}}\mathrm{h} \\ \frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2=\left(5.29-0.265\right)×19.4 \\ \frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2=98.09 \\ {\mathrm{v}}_{\mathrm{f}}^2=\frac{2×98.09}{5.29/9.8} \\ =363 \\ {\mathrm{v}}_{\mathrm{f}}=19.0\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the final speed of the stone before it comes to the ground = 19.0 m/s