Q5E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

An object of mass 5 kg is given an initial downward velocity of 50 m/sec and then allowed to fall under the influence of gravity. Assume that the force in newtons due to air resistance is -10v, where v is the velocity of the object in m/sec. Determine the equation of motion of the object. If the object is initially 500 m above the ground, determine when the object will strike the ground.

Step-by-Step Solution

Answer

• The equation of motion of the object is $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{91}\mathbf{t}\mathbf{+}\mathbf{22}\mathbf{.}\mathbf{55}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{)}$
• The time takes the object hits the ground 97.24 sec.

Step 1: Find the velocity

 $$\begin{gathered} \text{ma = W - bv}\left(\text{t}\right) \\ \text{ma = mg - bv}\left(\text{t}\right) \end{gathered}$$

Velocity,

$$\begin{gathered} \mathrm{v}\left(\mathrm{t}\right)=\frac{\mathrm{mg}}{\mathrm{b}}+(\mathrm{v}-\frac{\mathrm{mg}}{\mathrm{b}}\left){\mathrm{e}}^{\frac{-\mathrm{bt}}{\mathrm{m}}} \\ \mathrm{v}\left(\mathrm{t}\right)=\frac{5(9.81)}{10}+(50-\frac{5(9.81)}{10}\right){\mathrm{e}}^{\frac{-10\mathrm{t}}{5}} \\ \mathrm{v}\left(\mathrm{t}\right)=4.905+45.095{\mathrm{e}}^{-2\mathrm{t}} \end{gathered}$$

Step 2: Find the equation of motion

$$\begin{gathered} \mathrm{x}\left(\mathrm{t}\right)=\frac{\mathrm{mgt}}{\mathrm{b}}+\frac{\mathrm{m}}{\mathrm{b}}(\mathrm{v}-\frac{\mathrm{mg}}{\mathrm{b}}\left)\right(1-{\mathrm{e}}^{\frac{-\mathrm{bt}}{\mathrm{m}}}) \\ \mathrm{x}\left(\mathrm{t}\right)=4.905\mathrm{t}+\frac{5}{10}(45.095\left)\right(1-{\mathrm{e}}^{-2\mathrm{t}}) \\ =4.91\mathrm{t}+22.55(1-{\mathrm{e}}^{-2\mathrm{t}}) \end{gathered}$$

Hence the equation of motion is $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{91}\mathbf{t}\mathbf{+}\mathbf{22}\mathbf{.}\mathbf{55}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{)}$

Step 3: Find the value of t

Put x=500 and neglecting the exponential part

 

$$\begin{gathered} 500=4.91\mathrm{t}+22.55 \\ \mathrm{t}=97.24 \sec \end{gathered}$$

Hence, the time takes the object hits the ground 97.24 sec.