# B Average Passenger Plane Terminal velocity with added speed

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1. May 12, 2016

A passenger plane is traveling at a speed of 500 knots, has a mass of 72574.7792 kg, and has an altitude of 35,000 feet. If the pilot lost control of the plane and couldn't reduce speed or anything and the plane was going down at approximately a 40 degree angle, how long would you have until it hit the ground?

This is just curiousity btw, so if it is too hard to find out, then don't worry about it

2. May 12, 2016

### Staff: Mentor

Welcome to the PF.

Have you had trigonometry yet in school? That's the easiest way to figure out this problem.

3. May 12, 2016

No i have not had it yet

4. May 12, 2016

Only if you neglect air resistance. Given that the title included the phrase "terminal velocity", that would imply that the question is not intended to neglect air resistance. In that case, this is not a simple problem to solve. You'd have to simply estimate to some arbitrary (and likely poor) level of accuracy.

5. May 12, 2016

### FactChecker

That is a pretty steep decent so I assume the plane's speed would go higher than a normal cruise speed or any safe speed. The terminal speed is not something that would normally be studied or published, so you will have to make some assumptions. And of course, the terminal speed is the main thing that would determine the time before the crash.

6. May 12, 2016

### jbriggs444

That is quoting the mass to the precision of 1/10 of a gram. You would have been better off to say 160,000 pounds.

7. May 12, 2016

### Staff: Mentor

True, but I assumed that the speed would not be much more than 500 knots. Maybe that was too much of a simplification on my part.

8. May 13, 2016

### A.T.

Then download X-Plane and try it out. Much more fun than trying to calculate something like this.
http://www.x-plane.com/desktop/home/

9. May 13, 2016

Regardless of speed, air resistance complicated matters substantially. If there was no resistance then it is sime kinematic a. If it was a sphere with air resistance, that's relatively easy to estimate. But with a plane, it will depend so much on what orientation the plane has as it is falling that it's an intractable problem.

10. May 13, 2016

### CWatters

Looks like an insect sneaked on board that day and pushed the weight up an extra 0.2g. :-)

11. May 13, 2016

### Khashishi

Clearly, we need to make some gross approximations.
Here is a table of some typical drag coefficients. http://www.engineeringtoolbox.com/drag-coefficient-d_627.html
If we take the plane to have a drag coefficient of $c_d = 0.012$, and a frontal area of $A = 10 m^2$, and density of air $\rho = 1.2 kg/m^3$,
$F_d = c_d \rho v^2 A$
Terminal velocity $v_f$ is reached when the vertical component of the drag equals the force of gravity.
$mg = F_d \sin 40^o$
therefore
$v_f = \sqrt{ \frac{mg}{c_d \rho A \sin 40^o}}$
~ 2800 m/s
Is it really that high? Maybe the drag is higher than 0.012.

Anyways, you probably wouldn't have enough time to reach terminal velocity.