Irodov 1.104: Drag Costs More Going Up – Amit & Deepali Sinha Foundation

Problem Irodov 1.104

A body of mass $m$ is thrown vertically upward with initial speed $v_0$ . Air drag on the body has magnitude $kv^2$ at any instant its speed is $v$ , for some constant $k$ , and always opposes the body’s velocity. Find the speed $v'$ with which the body returns to its starting height.

Given

$m$
$v_0$
drag $=kv^2$ , always opposing $\vect v$

Find

The return speed

v'

at the starting height

Solution

Step 1Equation of motion on the way up

Rising: gravity m g and the drag kv^2 both point downward, since drag opposes the upward velocity v (drawn with a small angular offset from m g purely so the two are visible as separate arrows). — Rising: gravity $m\vect g$ and the drag $kv^2$ both point downward, since drag opposes the upward velocity $\vect v$ (drawn with a small angular offset from $m\vect g$ purely so the two are visible as separate arrows).

Take $y$ positive upward from the launch point. While rising, both gravity and drag point downward (figure above), so with $v=dy/dt>0$ ,

m\frac{dv}{dt} = -mg-kv^2.

Using $dv/dt=v\,dv/dy$ and separating variables,

\frac{v\,dv}{g+(k/m)v^2} = -dy.

Step 2The height reached

Integrate from the launch point $y=0$ , $v=v_0$ , to the top $y=H$ , $v=0$ :

\int_{v_0}^{0}\frac{v\,dv}{g+(k/m)v^2} = -\int_0^H dy.

The left side is a logarithm of the denominator, evaluated at its two limits:

\frac{m}{2k}\left[\ln\!\left(g+\frac{k}{m}v^2\right)\right]_{v_0}^{0} = -H \quad\Longrightarrow\quad \frac{m}{2k}\ln\frac{g}{g+(k/m)v_0^2} = -H.

H = \dfrac{m}{2k}\ln\!\left(1+\dfrac{kv_0^2}{mg}\right)

(1)

Step 3Equation of motion on the way down

Falling: gravity m g still points downward, while the drag kv^2, opposing the now downward velocity v, points up. Velocity and drag are drawn slightly off vertical so they read as arrows separate from m g. — Falling: gravity $m\vect g$ still points downward, while the drag $kv^2$ , opposing the now downward velocity $\vect v$ , points up. Velocity and drag are drawn slightly off vertical so they read as arrows separate from $m\vect g$ .

Take $y'$ positive downward from the top, so $y'=0$ is the top and $y'=H$ is the launch height. While falling, gravity points down (in the $+y'$ sense) and drag opposes the downward velocity, so it points up (figure above):

m\frac{dv}{dt} = mg-kv^2.

Rewriting the acceleration as $v\,dv/dy'$ again and separating the variables,

\frac{v\,dv}{g-(k/m)v^2} = dy'.

Step 4The return speed

Integrate from the top $y'=0$ , $v=0$ , to the launch height $y'=H$ , $v=v'$ :

\int_0^{v'}\frac{v\,dv}{g-(k/m)v^2} = \int_0^{H} dy' \quad\Longrightarrow\quad -\frac{m}{2k}\ln\frac{g-(k/m)v'^2}{g} = H.

Substituting (1) for $H$ and canceling the common factor $m/(2k)$ ,

-\ln\frac{g-(k/m)v'^2}{g} = \ln\!\left(1+\frac{kv_0^2}{mg}\right) \quad\Longrightarrow\quad \frac{g}{g-(k/m)v'^2} = 1+\frac{kv_0^2}{mg}.

Cross multiplying and solving for $v'^2$ ,

\begin{align*}g &= \left(g-\frac{k}{m}v'^2\right)\left(1+\frac{kv_0^2}{mg}\right) \\ g\cdot\frac{mg}{mg+kv_0^2} &= g-\frac{k}{m}v'^2 \\ \frac{k}{m}v'^2 &= g-\frac{mg^2}{mg+kv_0^2} = \frac{gkv_0^2}{mg+kv_0^2}\end{align*}

v'^2 = \frac{gv_0^2}{mg+kv_0^2}\cdot m = \frac{v_0^2}{1+kv_0^2/(mg)}

v' = \dfrac{v_0}{\sqrt{1+kv_0^2/(mg)}}

(2)

Check

As $k\to0$ (no drag), $v'\to v_0$ , recovering ordinary free flight where the body returns at launch speed. Also $v'<v_0$ for any $k>0$ , as it must: drag removes mechanical energy on both the way up and the way down, so the body cannot return faster than it left.

Remark

The ratio $v'^2/v_0^2=1/(1+kv_0^2/mg)$ depends on $k$ , $v_0$ , $m$ and $g$ only through the single dimensionless combination $kv_0^2/(mg)$ , even though the intermediate height $H$ needed its own logarithm to reach. That collapse happens because integrating $v\,dv$ against either $g+(k/m)v^2$ (Step 2) or $g-(k/m)v^2$ (Step 4) produces a logarithm with the exact same prefactor $m/(2k)$ out front; setting the two results equal through the common height $H$ cancels that shared prefactor, leaving only the logarithms’ arguments to combine.