Falling with Drag

Drop something and gravity pulls it down with constant force mg, but air resistance grows as it speeds up. With drag modeled as kv, opposing motion, Newton's second law reads m dv/dt = mg - kv. At the instant of release v = 0, so drag does nothing and acceleration is a full g. As v grows, kv eats into the net force, so acceleration shrinks. Acceleration never flips sign or hits zero in finite time, it just decays toward zero, so velocity creeps up toward the terminal value v_term = mg/k and gets arbitrarily close without ever quite touching it. Solving the differential equation gives v(t) = (mg/k)(1 - e^−kt/m), an exponential rise.

The speed at which this approach happens is set by the time constant m/k. A coffee filter is light and very draggy, so m/k is tiny, it reaches its low terminal velocity almost the instant it is released. A rock is heavy and streamlined, so m/k is huge and its terminal velocity itself is enormous, it would need to fall for a very long time to get anywhere near it, so in a normal fall off a building it just keeps speeding up at close to g. Graph the motion: velocity rises and flattens toward v_term, acceleration starts at g and decays exponentially toward zero, and position curves from a parabola-like start into a straight line of slope v_term once terminal velocity is essentially reached.

This topic is where Unit 2's differential equations become real: instead of constant acceleration you solve m dv/dt = mg - kv directly, using separation of variables, the same method used for other drag and resistive-force problems. If an object is thrown downward faster than terminal speed, drag exceeds gravity, so kv > mg and the net force points upward, decelerating the object back down toward v_term from above instead of up toward it.