Time to Reach Maximum Height for Particle Thrown Upward With Air Resistance Proportional to Speed Squared
\[m\]
is proportional to the square of the speed, so that \[R=kv^2\]
/ 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 \[v_0=\frac{dy}{dt}\|_{t=0}\]
. Applying \[F=ma\]
to the particle while it is moving upwards gives\[ma =-mg-kv^2\]
Using
\[v= \frac{dy}{dt}, \; a= \frac{dv}{dt}\]
this equation becomes \[m \frac{dv}{dt} = -mg - kv^2 \rightarrow \frac{m}{mg+kv^2}\frac{dv}{dt} =-1\]
.substitute
\[v= \sqrt{ \frac{mg}{k}} tan \theta\]
then \[dv = \sqrt{ \frac{mg}{k}} sec^2 \theta \]
, The equation becomes \[\sqrt{ \frac{m}{gk}} d \theta = -dt\]
.Integration gives
\[\sqrt{ \frac{m}{gk}} \theta=-t+C \rightarrow \sqrt{ \frac{m}{gk}} tan^{-1} (\sqrt{k}{mg}} v)=-t+C\]
.When
\[t=0, v=v_0 \rightarrow C=\sqrt{ \frac{m}{gk}} tan^{-1} (\sqrt{k}{mg}} v_0)\]
. 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
\[v=0\]
, at \[t= \sqrt{ \frac{m}{gk}} tan^{-1} (\sqrt{k}{mg}} v_0)\]
. 
