Finding Fall Time With Terminal Velocity

Question

I am calculating the fall time of an object $\frac{gt^2}{2} + vt + y = \beta$ where:

• $g$ is -32
• $v$ is 1
• $y$ is 500
• $\beta$ is -1000

Since I only want positive time I'll only consider the addition component of the quadratic equation, so about 9.7.

Now I want to add terminal velocity, so I need to find the point at which the object has reached terminal velocity, but I think my equation is bad. I'm solving $\frac{gt^2}{2} + vt = \Omega$ where:

• $g$ is -32
• $v$ is 1
• $\Omega$ is -320

Again taking only the addition component of the quadratic equation I get about 4.5.

So now I would expect that I need to solve for the object height at time 4.5 in the original equation so about 339.4. So I need to find the remaining fall distance which will be about 1160.6. And find the time at terminal velocity to travel that distance so about 3.6 seconds.

Now the problem, the time to terminal velocity is 4.5 seconds, the time from the height at which terminal velocity is reached to the ground is 3.6 seconds so the total fall time with terminal velocity is 8.1 seconds.

Meaning it will hit the ground faster when the terminal velocity calculation is applied. So I've done something wrong. Can someone point out what my error is?

Answer 1

The equation: $\frac{gt^2}{2} + vt = \Omega$ is incorrect.

To find the right equation think about the units. $g$ is in $\frac{\mathrm{feet}}{\mathrm{second}^2}$ and $v$ is in $\frac{\mathrm{feet}}{\mathrm{second}}$. So the left hand side of the equation will give you a result in feet. So when you plugged your numbers into this equation you solved for how many seconds it would take the object to fall 320 feet.

You're trying to find the time at which the acceleration has pushed the object to a given speed, so $\frac{\mathrm{feet}}{\mathrm{second}}$. To find that you need to sum the initial velocity and how many seconds of acceleration you have been under. So your equation should be:

$$\frac{gt}{2} + v = \Omega$$

Plugging your numbers back in you'll find that will take 20.0625 seconds, so the object will have hit the ground before it reaches terminal velocity as it only fell about 9.7 seconds before considering terminal velocity.

Answer 2

At terminal velocity, there is no acceleration. You don't identify the variables in the second equation, but to compute the terminal velocity you should have $mg=\frac 12c_dA\rho v^2$ where $c_d$ is the drag coefficient, $A$ the frontal area, $\rho$ the density of air, and $v$ the terminal velocity. This shows the balance of forces-gravity down, air resistance up-resulting in no acceleration.

Answer 3

Since you haven't explained the reasoning, it's hard to explain where it has gone wrong. In particular, what is $\Omega$? If it is intended to be the terminal velocity, why should it equal the distance $gt^2/2+vt$ traversed after time $t$ of free fall?

I suspect the problem is that you are using the equation for free fall, whereas terminal velocity is for fall with air resistance which follows a different equation. There is no point at which the terminal velocity is reached, but as the velocity gets closer to it the air resistance gets closer to the gravitational force making the acceleration slow down.

The acceleration of an object subject to gravitation and air resistance (at moderate speeds at least) is on the form $a=g-\alpha v^2$ where $\alpha$ is a constant (or fairly constant) which depends on the properties of the body and the air (see answer from Ross Millikan or Wikipedia on drag). Terminal velocity is reached when $a=0$: i.e., at $\Omega=\sqrt{g/\alpha}$

Solving the differential equation $a=dv/dt=g-\alpha v^2$ assuming $v=0$ at $t=0$ gives $$ v =\sqrt{\frac{g}{\alpha}}\tanh\left(\sqrt{\alpha g}\cdot t\right) =\Omega\cdot\tanh\left(\frac{gt}{\Omega}\right). $$