Falling Bodies

You already know about a falling body where the gravitational force is considered constant (which is valid if the body stays near the Earth's surface) and air resistance is neglected: the DE is y'' = - g where $g = 9.8\ m/s^2$, and y represents height off the ground. Now we will consider two complications:

1.

air resistance not neglected (on this page)

2.

gravitational force not constant (on next page).

Air resistance is a force in the opposite direction to the velocity. Its magnitude depends on the speed. The actual function can be rather complicated, but at low speeds we can assume that it is proportional to the speed. We can then write the differential equation as

my'' = -mg - k y'

where k is a positive constant. This is a second order equation, but for the velocity v = y' we have a first order equation:

m v' = -m g - k v

The general solution of this first order linear equation is

$$v = -\frac{mg}{k} + C {\rm e}^{-k t/m}$$

Note that as $t \to \infty$ the exponential goes to 0 so that $v \to -mg/k$. This is called the terminal velocity. At the terminal velocity, the forces of air resistance and gravity are exactly in balance, so there is no net acceleration. The terminal velocity is a stable equilibrium point of the differential equation for v.

To obtain the height y as a function of t, we can integrate v:

$$y = A - \frac{mg}{k} t - \frac{C m}{k} {\rm e}^{-k t/m}$$

Suppose we throw a ball upward from the surface of the earth with initial speed v₀. Thus the initial conditions are y(0) = 0 and v(0) = v₀. We have C = v₀ + mg/k and A = Cm/k, so

$$y = \frac{m}{k} \left(v_0 + \frac{mg}{k}\right)\left(1 - {\rm e}^{-k t/m}\right) - \frac{mg}{k}t$$

How long will the ball be in the air? To answer this we would have to solve the equation

$$\frac{m}{k} \left(v_0 + \frac{mg}{k}\right)\left(1 - {\rm e}^{-k t/m}\right) - \frac{mg}{k}t = 0$$

which can't be done in ``closed form''. Numerical methods such as Newton's would have to be used. But we can turn the question around, and ask what v₀ must be to keep the ball in the air for a given time T. The answer is

$$v_0 = \frac{g T}{1- {\rm e}^{-kT/m}} - \frac{mg}{k}$$

Example: A baseball has a mass of 0.145 kg, and k = .03 kg/s. In a ``pop fly'', it is hit straight up into the air, and comes back down exactly 7 seconds later. What was its initial speed, and what height did it reach?

Answer: 42.3 m/sec, 58.35 metres. This would hit the roof in many domed stadiums (including Montreal's Olympic Stadium), but not in Toronto's Skydome. I don't know about BC Place.