## 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
$m$
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
$t= \sqrt{ \frac{m}{gk}} tan^{-1} (\sqrt{k}{mg}} v_0)$
.

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