# Air resistance proportional to velocity squared

Write the equation of the velocity of a body falling, if air resistance is proportional to the velocity squared. If $$g=32$$ ft/sec$$^2$$, and the constant of proportionality is $$c=0.25,$$ consider the initial velocity as $$v(0) =20$$ ft/sec.

1. Set up the differential equation and solve it under the given initial value problem.
2. Calculate numerically and graph the velocity in the interval [0, 10].
3. Estimate the terminal velocity in this case.

I found an equation that may help:

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

But is this is a correct formula to solve the problem, how can I solve it if I don't have the value of m?

• Divide this equation by m. You'll be left with a ${k \over m}$ factor, which is presumably equal to the $c=0.25$ from the problem statement. Commented Feb 26, 2023 at 21:41
• I would argue that the problem as stated is unsolvable, because "air resistance" is a force, so $cv^2$ must have units of force and cannot be compared with an acceleration. But I think the interpretation in the previous comment is probably what was intended. Commented Feb 26, 2023 at 21:44

Write the equation of the velocity of a body falling, if air resistance is proportional to the velocity squared. If $$g=32$$ ft/sec$$^2$$, and the constant of proportionality is $$c=0.25,$$ consider the initial velocity as $$v(0) =20$$ ft/sec.
As pointed out by David, the given exercise is not dimensionally consistent: the air resistance should have been specified as $$0.25\frac{\text{kg}}{\text{m}}\cdot v^2$$ instead of $$0.25v^2$$ (in this quadratic model of air resistance, the coefficient encodes the body's shape and size and the surrounding air's viscosity and density) or, better, rephrased the exercise to make clear that all variables represent their corresponding numeric values (eg., given the acceleration, instead of $$g,$$ as $$32$$ ft/sec$$^2$$). If we do this, then, since the net downward force acting on the body equals its weight minus its air resistance, we have that $$m\frac{dv}{dt}=mg-0.25v^2.$$ This equation (not "formula", btw) can be solved by treating $$g$$ and $$m$$ as parameters/constants.
You're right that without $$m$$'s value, the answers to parts 2 and 3 of this exercise can't be determined.