Answer on Question#38807 - Physics, Mechanics | Kinematics | Dynamics
A Parachutist is falling with a speed of 50 m / s 50\mathrm{m/s} 50 m/s ( M v ∧ 2 ) / 25 (\mathrm{Mv}^{\wedge}2)/25 ( Mv ∧ 2 ) /25 g = 10 m / s g=10\mathrm{m/s} g = 10 m/s
Solution:
Taking downward to be positive, we have:
d v d t = F m = M g − M v 2 25 M = g − v 2 25 g = 10 m s 2 ⇒ d v d t = 1 25 ( 250 − v 2 ) \begin{array}{l}
\frac{\mathrm{d}v}{\mathrm{d}t} = \frac{F}{m} = \frac{Mg - \frac{Mv^2}{25}}{M} = g - \frac{v^2}{25} \\
g = 10 \frac{m}{s^2} \Rightarrow \\
\frac{dv}{dt} = \frac{1}{25} (250 - v^2)
\end{array} d t d v = m F = M M g − 25 M v 2 = g − 25 v 2 g = 10 s 2 m ⇒ d t d v = 25 1 ( 250 − v 2 )
Divide both sides by 1 25 ( 250 − v 2 ) \frac{1}{25} (250 - v^2) 25 1 ( 250 − v 2 )
25 d v d t − v 2 + 250 = 1 \frac{25 \frac{dv}{dt}}{-v^2 + 250} = 1 − v 2 + 250 25 d t d v = 1
Integrate both sides with respect to x:
∫ 25 d v d t − v 2 + 250 d t = ∫ 1 d t \int \frac{25 \frac{dv}{dt}}{-v^2 + 250} dt = \int 1 dt ∫ − v 2 + 250 25 d t d v d t = ∫ 1 d t
Evaluate the integrals:
− 1 2 ( 5 2 ( log ( − v + 5 10 ) − log ( v + 5 10 ) ) ) = t + c 1 , where c 1 is arbitrary constant \begin{array}{l}
- \frac{1}{2} \left(\sqrt{\frac{5}{2}} \left(\log(-v + 5\sqrt{10}) - \log(v + 5\sqrt{10})\right)\right) \\
= t + c_1, \text{ where } c_1 \text{ is arbitrary constant}
\end{array} − 2 1 ( 2 5 ( log ( − v + 5 10 ) − log ( v + 5 10 ) ) ) = t + c 1 , where c 1 is arbitrary constant
Solve for v:
v ( t ) = 5 10 ( e 2 2 5 ( t + c 1 ) − 1 ) e 2 2 5 ( t + c 1 ) + 1 v(t) = \frac{5\sqrt{10} \left(e^{2\sqrt{\frac{2}{5}} (t + c_1)} - 1\right)}{e^{2\sqrt{\frac{2}{5}} (t + c_1)} + 1} v ( t ) = e 2 5 2 ( t + c 1 ) + 1 5 10 ( e 2 5 2 ( t + c 1 ) − 1 )
Solve for c 1 c_1 c 1
Substitute v ( 0 ) = 50 m s v(0) = 50 \frac{m}{s} v ( 0 ) = 50 s m v ( t ) = 5 10 ( e 2 2 5 ( t + c 1 ) − 1 ) e 2 2 5 ( t + c 1 ) + 1 v(t) = \frac{5\sqrt{10} \left(e^{2\sqrt{\frac{2}{5}} (t + c_1)} - 1\right)}{e^{2\sqrt{\frac{2}{5}} (t + c_1)} + 1} v ( t ) = e 2 5 2 ( t + c 1 ) + 1 5 10 ( e 2 5 2 ( t + c 1 ) − 1 )
5 10 ( e 2 2 5 c 1 − 1 ) e 2 2 5 c 1 + 1 = 50 \frac{5\sqrt{10} \left(e^{2\sqrt{\frac{2}{5}} c_1} - 1\right)}{e^{2\sqrt{\frac{2}{5}} c_1} + 1} = 50 e 2 5 2 c 1 + 1 5 10 ( e 2 5 2 c 1 − 1 ) = 50
Solve the equation:
c 1 = 5 2 log ( 1 3 ( − i ) 11 + 2 10 ) c_1 = \sqrt{\frac{5}{2}} \log \left(\frac{1}{3}(-i)\sqrt{11 + 2\sqrt{10}}\right) c 1 = 2 5 log ( 3 1 ( − i ) 11 + 2 10 ) c 1 = 5 2 log ( 1 3 i 11 + 2 10 ) c_1 = \sqrt{\frac{5}{2}} \log \left(\frac{1}{3}i\sqrt{11 + 2\sqrt{10}}\right) c 1 = 2 5 log ( 3 1 i 11 + 2 10 )
Substitute c 1 c_1 c 1 v ( t ) = 5 10 ( e 2 2 5 ( t + c 1 ) − 1 ) e 2 2 5 ( t + c 1 ) + 1 v(t) = \frac{5\sqrt{10}\left(e^{2\sqrt{\frac{2}{5}(t + c_1)}}-1\right)}{e^{2\sqrt{\frac{2}{5}(t + c_1)}}+1} v ( t ) = e 2 5 2 ( t + c 1 ) + 1 5 10 ( e 2 5 2 ( t + c 1 ) − 1 )
v ( t ) = 5 10 ( exp ( 2 2 5 ( t + 5 2 log ( 1 3 i 11 + 2 10 ) ) ) − 1 ) exp ( 2 2 5 ( t + 5 2 log ( 1 3 i 11 + 2 10 ) ) ) + 1 v(t) = \frac{5\sqrt{10}\left(\exp\left(2\sqrt{\frac{2}{5}}\left(t + \sqrt{\frac{5}{2}}\log\left(\frac{1}{3}i\sqrt{11 + 2\sqrt{10}}\right)\right)\right) - 1\right)}{\exp\left(2\sqrt{\frac{2}{5}}\left(t + \sqrt{\frac{5}{2}}\log\left(\frac{1}{3}i\sqrt{11 + 2\sqrt{10}}\right)\right)\right) + 1} v ( t ) = exp ( 2 5 2 ( t + 2 5 log ( 3 1 i 11 + 2 10 ) ) ) + 1 5 10 ( exp ( 2 5 2 ( t + 2 5 log ( 3 1 i 11 + 2 10 ) ) ) − 1 )
Alternate form:
v ( t ) = − 55 10 e 2 2 5 t 9 ( 1 − 1 9 ( 11 + 2 10 ) e 2 2 5 t ) − 100 e 2 2 5 t 9 ( 1 − 1 9 ( 11 + 2 10 ) e 2 2 5 t ) − 5 10 1 − 1 9 ( 11 + 2 10 ) e 2 2 5 t v(t) = - \frac{55\sqrt{10}e^{2\sqrt{\frac{2}{5}}t}}{9\left(1 - \frac{1}{9}(11 + 2\sqrt{10})e^{2\sqrt{\frac{2}{5}}t}\right)} - \frac{100e^{2\sqrt{\frac{2}{5}}t}}{9\left(1 - \frac{1}{9}(11 + 2\sqrt{10})e^{2\sqrt{\frac{2}{5}}t}\right)} - \frac{5\sqrt{10}}{1 - \frac{1}{9}(11 + 2\sqrt{10})e^{2\sqrt{\frac{2}{5}}t}} v ( t ) = − 9 ( 1 − 9 1 ( 11 + 2 10 ) e 2 5 2 t ) 55 10 e 2 5 2 t − 9 ( 1 − 9 1 ( 11 + 2 10 ) e 2 5 2 t ) 100 e 2 5 2 t − 1 − 9 1 ( 11 + 2 10 ) e 2 5 2 t 5 10
Approximated expanded form:
v ( t ) = 37 e 2.5 t − 19 e 1.26 t + 11 e 1.26 t − 21 e 2.5 t + 1422 173 e 1.26 t − 90 = 167 + 1422 610 t − 90 v(t) = 37e^{2.5t} - 19e^{1.26t} + 11e^{1.26t} - 21e^{2.5t} + \frac{1422}{173e^{1.26t} - 90} = 167 + \frac{1422}{610t - 90} v ( t ) = 37 e 2.5 t − 19 e 1.26 t + 11 e 1.26 t − 21 e 2.5 t + 173 e 1.26 t − 90 1422 = 167 + 610 t − 90 1422
Answer: speed of the man as a function of time t t t
v ( t ) = 167 + 1422 610 t − 90 . v(t) = 167 + \frac{1422}{610t - 90}. v ( t ) = 167 + 610 t − 90 1422 . 
