Step 1

We have given that the mass of parachutist is $m$.

We have given that the initial velocity of the parachutist is zero.

We are given that the drag resistance force ${F_D}$ acting on parachutist is ${F_D} = k{v^2}$, where, k is a constant and v is the velocity.

We are asked to calculate the velocity of the parachutist when it reaches the ground.

 
Step 2

Draw a free-body diagram of the parachutist.

Here, $a$ is the acceleration of the parachutist in the vertically downward direction.

 
Step 3

Apply equilibrium equation of motion in the vertical direction:

\[\begin{array}{c} \sum {{F_y}} = 0\\ mg - {F_D} = ma \end{array}\]

Substitute the given value of ${F_D}$ in the above equation:

\[\begin{array}{c} mg - k{v^2} = m\frac{{dv}}{{dt}}\\ dt = \frac{{dv}}{{g - \left( {\frac{{k{v^2}}}{m}} \right)}}\\ dt = \frac{m}{k}\left( {\frac{{dv}}{{\frac{{mg}}{k} - {v^2}}}} \right) \end{array}\]……(1)
 
Step 4

Integrate equation (1) to get the equation for velocity:

\[\begin{array}{c} \int\limits_0^t {dt} = \int\limits_0^v {\frac{m}{k}\left( {\frac{{dv}}{{\frac{{mg}}{k} - {v^2}}}} \right)} \\ \left[ t \right]_0^t = \frac{m}{k}\left[ {\left( {\frac{1}{{2\sqrt {\frac{{mg}}{k}} }}} \right)\ln \left( {\frac{{\sqrt {\frac{{mg}}{k}} + v}}{{\sqrt {\frac{{mg}}{k}} - v}}} \right)} \right]_0^v\\ \left( {t - 0} \right) = \frac{m}{k}\left( {\frac{1}{{2\sqrt {\frac{{mg}}{k}} }}} \right)\left[ {\ln \left( {\frac{{\sqrt {\frac{{mg}}{k}} + v}}{{\sqrt {\frac{{mg}}{k}} - v}}} \right) - \ln \left( {\frac{{\sqrt {\frac{{mg}}{k}} + 0}}{{\sqrt {\frac{{mg}}{k}} - 0}}} \right)} \right]\\ t = \frac{m}{k}\left( {\frac{1}{{2\sqrt {\frac{{mg}}{k}} }}} \right)\ln \left( {\frac{{\sqrt {\frac{{mg}}{k}} + v}}{{\sqrt {\frac{{mg}}{k}} - v}}} \right) \end{array}\]

Further, simplify the above equation for velocity:

\[\begin{array}{c} \ln \left( {\frac{{\sqrt {\frac{{mg}}{k}} + v}}{{\sqrt {\frac{{mg}}{k}} - v}}} \right) = \left( {2\sqrt {\frac{{mg}}{k}} } \right)\left( {\frac{k}{m}} \right)t\\ \ln \left( {\frac{{\sqrt {\frac{{mg}}{k}} + v}}{{\sqrt {\frac{{mg}}{k}} - v}}} \right) = \left( {2\sqrt {\frac{{kg}}{m}} } \right)t\\ {e^{2t\sqrt {\frac{{kg}}{m}} }} = \frac{{\sqrt {\frac{{mg}}{k}} + v}}{{\sqrt {\frac{{mg}}{k}} - v}}\\ \sqrt {\frac{{kg}}{m}} {e^{2t\sqrt {\frac{{mg}}{k}} }} - v{e^{2t\sqrt {\frac{{mg}}{k}} }} = \sqrt {\frac{{mg}}{k}} + v \end{array}\]

Solving further the above equation:

\[\begin{array}{c} \sqrt {\frac{{kg}}{m}} {e^{2t\sqrt {\frac{{mg}}{k}} }} - \sqrt {\frac{{mg}}{k}} = v\left( {1 + {e^{2t\sqrt {\frac{{mg}}{k}} }}} \right)\\ v = \sqrt {\frac{{kg}}{m}} \left( {\frac{{{e^{2t\sqrt {\frac{{mg}}{k}} }} - 1}}{{{e^{2t\sqrt {\frac{{mg}}{k}} }} + 1}}} \right) \end{array}\]……(2)

Equation (2) shows the velocity of the parachutist when he lands on the ground.

 
Step 5

The equation (2) can be written as:

\[\begin{array}{c} v = \sqrt {\frac{{kg}}{m}} \left( {\frac{{{e^{2t\sqrt {\frac{{mg}}{k}} }}}}{{{e^{2t\sqrt {\frac{{mg}}{k}} }}}}} \right)\left( {\frac{{1 - \frac{1}{{{e^{2t\sqrt {\frac{{mg}}{k}} }}}}}}{{1 + \frac{1}{{{e^{2t\sqrt {\frac{{mg}}{k}} }}}}}}} \right)\\ v = \sqrt {\frac{{kg}}{m}} \left( {\frac{{1 - {e^{ - 2\sqrt {\frac{{mg}}{k}} t}}}}{{1 + {e^{ - 2\sqrt {\frac{{mg}}{k}} t}}}}} \right) \end{array}\]
 
Step 6

Calculate the terminal velocity ${v_\infty }$ at an infinite time by substituting $t = \infty $ in the above equation:

\[\begin{array}{c} {v_\infty } = \sqrt {\frac{{kg}}{m}} \left( {\frac{{1 - {e^{ - 2\sqrt {\frac{{mg}}{k}} \left( \infty \right)}}}}{{1 + {e^{ - 2\sqrt {\frac{{mg}}{k}} \left( \infty \right)}}}}} \right)\\ {v_\infty } = \sqrt {\frac{{kg}}{m}} \left( {\frac{{1 - {e^{ - \infty }}}}{{1 + {e^{ - \infty }}}}} \right)\\ {v_\infty } = \sqrt {\frac{{kg}}{m}} \left( {\frac{{1 - 0}}{{1 + 0}}} \right)\\ {v_\infty } = \sqrt {\frac{{kg}}{m}} \end{array}\]