Average Passenger Plane Terminal velocity with added speed

[ThePizzaDeliveryGuy]

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

 

[berkeman]

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

Welcome to the PF.

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

 

[ThePizzaDeliveryGuy]

Welcome to the PF.

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

No i have not had it yet

 

[boneh3ad]

Welcome to the PF.

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

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.

 

[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.

 

[jbriggs444]

has a mass of 72574.7792 kg

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.

 

[berkeman]

In that case, this is not a simple problem to solve.

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.

 

[A.T.]

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

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/

 

[boneh3ad]

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.

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.

 

[CWatters]

A passenger plane is traveling at a speed of 500 knots, has a mass of 72574.7792 kg

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

 

[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.