Average Passenger Plane Terminal velocity with added speed

  • #1
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
 

Answers and Replies

  • #2
berkeman
Mentor
59,078
9,186
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
Welcome to the PF.

Have you had trigonometry yet in school? That's the easiest way to figure out this problem.
 
  • #3
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
 
  • #4
boneh3ad
Science Advisor
Insights Author
Gold Member
3,204
888
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.
 
  • #5
FactChecker
Science Advisor
Gold Member
6,057
2,341
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
jbriggs444
Science Advisor
Homework Helper
9,612
4,261
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.
 
  • #7
berkeman
Mentor
59,078
9,186
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.
 
  • #9
boneh3ad
Science Advisor
Insights Author
Gold Member
3,204
888
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.
 
  • #10
CWatters
Science Advisor
Homework Helper
Gold Member
10,532
2,298
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. :-)
 
  • Like
Likes FactChecker
  • #11
Khashishi
Science Advisor
2,815
493
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.
 
  • Like
Likes FactChecker

Related Threads on Average Passenger Plane Terminal velocity with added speed

Replies
6
Views
4K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
3
Views
11K
  • Last Post
Replies
11
Views
8K
  • Last Post
Replies
3
Views
17K
  • Last Post
Replies
4
Views
82K
  • Last Post
Replies
6
Views
7K
  • Last Post
Replies
6
Views
6K
Top