Differential Equation

When you integrated to find t you put the constant inside the log  term. If you'd simply added the constant to the right hand side you would have found that c₁ = 50ln(50g - 5k)/k and subsequently that k = 50g/65

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"Find the velocity of a falling body at t=5 seconds 

Air resistance is proportional to the velocity of the body.

weight of the body=50kg

Height = 100m

Initial velocity = 5m/sec

Limiting velocity = 65/sec"

I've assumed earth gravity above.  However, if you integrate again to find the height after 5 seconds you get a negative height, so perhaps this isn't on earth!  Also, air resistance is more normally taken to be proportional to velocity squared (though this means there is no nice closed form solution for velocity as a function of time - it has to be calculated numerically). 

Hi MarcoP,

I'll give it a shot, just don't rely on my answer :))

Find the velocity of a falling body at t=5 seconds 

weight of the body=50kg

Air resistance is proportional to the velocity of the body. 

weight of the body=50kg =m

  \(\begin{align}\\air\;\; resistance\;\; up \;&\;\alpha \;\; velocity \;\;down\\ mass*accel\;&\;\alpha \;\; velocity \;down \\ m\frac{dv}{dt}&=k*\frac{dx}{dt} \\ \frac{dv}{dt}&=\frac{k}{m}*\frac{dx}{dt} \end{align} \)

Height = 100m 

Initial velocity = 5m/sec

Limiting velocity = 65m/sec

Let downwards be positive velocity

\(Initially\\ \;\;t=0\;\;\;\;\  x=100 \;\;  \;\;  \dot x =5     \;\;  \;\;\ddot x=g-\frac{k}{50}*5=g-0.1k\)

Ongoing

\(\begin{align}\\\frac{dv}{dt}&=g-\frac{k}{50}*v\\ \frac{dv}{dt}&=\frac{50g-kv}{50}\\ \frac{dt}{dv}&=\frac{50}{50g-kv}\\ t&=\frac{50ln(50g-kv+c_1)}{-k}\\~\\ &When\;\;t=0 \;\;v=5\;\;find\;\; c_1\\~\\ 0&=\frac{50ln(50g-5k+c_1)}{-k}\\ 0&=50ln(50g-5k+c_1)\\ 1&=50g-5k+c_1\\ c_1&=1-50g+5k\\ \frac{-kt}{50}&=ln(50g-kv+1-50g+5k)\\ \frac{-kt}{50}&=ln(1-kv+5k)\\ e^{\frac{-kt}{50}}&=1-kv+5k\\ &As \;\;t\rightarrow\infty \quad v\rightarrow 65\\~\\ 0&=1-65k+5k\\ 0&=1-60k\\ 60k&=1\\ k&=\frac{1}{60}\\ so\\ e^{\frac{-t}{60*50}}&=1-\frac{v}{60}+\frac{5}{60}\\ e^{\frac{-t}{3000}}&=\frac{13}{12}-\frac{v}{60}\\ &When\;\; t=5\\ e^{\frac{-5}{3000}}&=\frac{13}{12}-\frac{v}{60}\\ e^{\frac{-1}{600}}&=\frac{65}{60}-\frac{v}{60}\\ 60e^{\frac{-1}{600}}&=65-v\\ v&=65-60e^{\frac{-1}{600}}\\ v&=5.1 \end{align}\\ \)

Which is wrong   :(

Alan can you see my error with any ease ?       

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Thanks Alan and Melody.

This will give me lots to chew on :)