Question

A stone with a weight of 52.9 Nis launched vertically from ground level with an initial speed of 20.0 m/s, and the air drag on it is 0.265 Nthroughout the flight. What are (a) the maximum height reached by the stone and (b) Its speed just before it hits the ground?

Short answer

a) The maximum height reached by the stone = 19.4 m

b) The final speed of the stone before it comes to the ground = 19.0 m/s

Step-by-step solution

Listing the given quantities

The weight of the stone = 52.9 N

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

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

Understanding the concept of law of conservation of energy

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:

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

Step 3(a): Calculation of maximum height reached by the stone

Using the conservation of energy principle,

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

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

Step 4(b): Calculation of final speed of the stone before it comes to the ground

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,

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

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