Question

A 30.0-kg object has a drag force with a magnitude proportional to the square of its speed. The object falls with an acceleration 4.00 m/s2 downward when it is falling downward at 70.0 m/s. What is its terminal velocity?

Answer 1

At terminal velocity $v_t$, the drag force has the same magnitude as the object's weight since the object is in equilibrium:

$F_{\rm net} = F_{\rm drag} - mg = 0 \implies F_{\rm drag} = (30.0\,\mathrm{kg})\left(9.80\dfrac{\rm m}{\mathrm s^2}\right) = 294\,\mathrm N$

Drag is proportional to the square of the speed, so there is some constant k such that

$F_{\rm drag} = kv^2$

and in particular, at terminal velocity we have

$294\,\mathrm N = k{v_t}^2$

At the moment the object is falling with speed v = 70.0 m/s and with acceleration a = 4.00 m/s², the net force on it is

$F_{\rm net}' = F_{\rm drag}' - mg = ma$

Solve for k :

$k\left(70.0\dfrac{\rm m}{\mathrm s^2}\right) - (30.0\,\mathrm{kg})\left(9.80\dfrac{\rm m}{\mathrm s^2}\right) = (30.0\,\mathrm{kg})\left(-4.00\dfrac{\rm m}{\mathrm s^2}\right) \\\\ \implies k \approx 0.0355 \dfrac{\rm kg}{\rm m}$

Now solve for the terminal velocity:

$294\,\mathrm N = \left(0.0355\dfrac{\rm kg}{\rm m}\right){v_t}^2 \implies \boxed{v_t \approx 91.0\dfrac{\rm m}{\rm s}}$