Time to Reach Maximum Height for Particle Thrown Upward With Air Resistance Proportional to Speed Squared

Suppose the air resistance on a particle of mass  
  is proportional to the square of the speed, so that  
/ A particle thrown up will then be subject to two forces slowing the particle - the forces of gravity and air resistance. Obviously when  
\[t=0, \; y=0\]
. Suppose the particle is thrown up with initial speed  
. Applying  
  to the particle while it is moving upwards gives
\[ma =-mg-kv^2\]

\[v= \frac{dy}{dt}, \; a= \frac{dv}{dt}\]
  this equation becomes  
\[m \frac{dv}{dt} = -mg - kv^2 \rightarrow \frac{m}{mg+kv^2}\frac{dv}{dt} =-1\]
\[v= \sqrt{ \frac{mg}{k}} tan \theta\]
\[dv = \sqrt{ \frac{mg}{k}} sec^2 \theta \]
, The equation becomes  
\[\sqrt{ \frac{m}{gk}} d \theta = -dt\]
Integration gives  
\[\sqrt{ \frac{m}{gk}} \theta=-t+C \rightarrow \sqrt{ \frac{m}{gk}} tan^{-1} (\sqrt{k}{mg}} v)=-t+C\]
\[t=0, v=v_0 \rightarrow C=\sqrt{ \frac{m}{gk}} tan^{-1} (\sqrt{k}{mg}} v_0)\]
. Then  
\[\sqrt{ \frac{m}{gk}} tan^{-1} (\sqrt{k}{mg}} v_0)- \sqrt{ \frac{m}{gk}} tan^{-1} (\sqrt{k}{mg}} v)=t\]
The greatest height is reached when  
, at  
\[t= \sqrt{ \frac{m}{gk}} tan^{-1} (\sqrt{k}{mg}} v_0)\]

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