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Maximum distance of flight

  • Thread starter roam
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  • #1
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Homework Statement



An airplane of mass 1280 kg has an engine failure when flying with an airspeed of 135 km/h at an altitude of 2590 m on a calm day. It then glides at a constant glide angle (which is the direction of flight below the horizontal) towards a safe landing at this constant speed of 135 km/h experiencing a drag force of 1310 N that opposes the direction in which the plane is moving.
Please use: g = 9.81 m s-2

Find the magnitude of the maximum distance over the ground the plane can glide while searching for a safe landing spot.

(Further info: the glide angle is 5.99, and the lift force which acts perpendicular to the wings of the plane is 12500).


The Attempt at a Solution



The correct solution must be 24700 m, can anyone show me how to get this answer?
I tried to first find the time the airplance has before it contacts the ground using the formula
[tex]y=v_{iy}t+\frac{1}{2}at^2[/tex]

[tex]v_{iy}=135 cos 5.9 = 134.2[/tex]

setting the formula equal to zero and solving for t

[tex]0=134.2t+\frac{1}{2}(-9.81)t^2[/tex]

t=0.0731

Then using the formula x=vt to find the maximum horizontal distance

[tex]135 \times 0.073 \neq 24700[/tex]

My method doesn't work...
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
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Resolve the drag force and lift force in to vertical and horizontal components. Then find the net horizontal and vertical components of the acceleration.
From the vertical component of the acceleration, find time to reach the ground. And from the horizontal component of the acceleration find the distance traveled by the plane before landing.
 
  • #3
1,266
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Resolve the drag force and lift force in to vertical and horizontal components. Then find the net horizontal and vertical components of the acceleration.
From the vertical component of the acceleration, find time to reach the ground. And from the horizontal component of the acceleration find the distance traveled by the plane before landing.
Finding the vertical and horizontal components of the drag force:

Fx=1310 cos 5.9 = 1303
Fy=1310 sin 5.9 = 134.6

horizontal and vertical components of the acceleration:

Fx=max=>1303 = 1280 ax
ax = 1.017

Fy=may=> 134.6 = 1280 ay
ay = 0.105

Is this alright so far? I don't undestand what to do to find the time from the vertical component. What formula should I use? I can't use v=vi+at because I end up with t=0. :grumpy:
 
  • #4
rl.bhat
Homework Helper
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Your ax and ay due to drag force is correct. Similarly find the x and y components of acceleration due to lift force. Then find Σax and Σay.
 
  • #5
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Your ax and ay due to drag force is correct. Similarly find the x and y components of acceleration due to lift force. Then find Σax and Σay.
I previously know that the lift force which acts perpendicular to the wings of the plane is 12500. So, is the following correct?

Finding the x and y components of acceleration due to lift force:

Fx = 12500 cos 5.9 = 12433.7
Fy = 12500 sin 5.9 = 1284.9

ax = 9.71
ay=1

If what I've done here is correct then

[tex]\sum a_x = 9.71 + 1.017 = 10.7[/tex]

and

[tex]\sum a_x = 1 + 0.105 = 1.105[/tex]
 
  • #6
rl.bhat
Homework Helper
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No.
Σay = 9.71 + 0.105 = 9.815 m/s^2
Σax = 1.017 - 1.003 =0.013 m/s^2
 
  • #7
1,266
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No.
Σay = 9.71 + 0.105 = 9.815 m/s^2
Σax = 1.017 - 1.003 =0.013 m/s^2
Thanks a lot for the correction :smile: but I still have some trouble with the calculation:

So to find the time from the ay I must use the formula y=viyt + 1/2 at2.

The airspeed 135 is in km/h, so I convert it to m/s because they want the final answer to be in m/s. Therefore v=37.5 m/s.

2590=37.5 t- 1/2 (9.819)t2

Solving the quadratic, I got t=3.82

Then using the formula x=vt, to find the distance I get 14.7 meters which is far less than 24700 m. What's the error?
 
  • #8
rl.bhat
Homework Helper
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2590=37.5 t- 1/2 (9.819)t2
This is not correct. Net acceleration in the y direction is
a = g - Σay.
The initial velocity is y component of the velocity of the plane.
Now find the time t.
 
  • #9
Filip Larsen
Gold Member
1,237
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You should probably notice, that since the flight path angle and the speed is constant during the glide, the velocity vector (i.e. both vertical and horizontal) is also constant which means the acceleration vector (both vertical and horizontal) is zero.

Zero acceleration means all forces acting on the (center of mass) of the plane must be balancing, and from analysis of this (and knowledge of the condition before the plane started gliding) you can calculate the glide path angle, if you want to.

With the glide path angle is given (either because you calculated it or because you used the value specified), you can now analyze how horizontal, vertical and total speed is related to this angle (hint: identify which side of a right-hand triangle that corresponds to which speed and which angle of the triangle that represent flight path angle and use trigonometry to related speeds to the flight path angle). Knowing the vertical speed you can now calculate the time it takes the plane to descend to zero altitude and from this time and the horizontal speed you can finally find the maximum range.
 
  • #10
1,266
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Ah, I see what you mean now by the net y acceleration! ay, net=(1280 × 9.81) -9.815 = 12546.98 ms2

y component of the velocity:

viy=> (135/3.6)m/s cos 5.9 = 37.3

37.3t - 1/2 × 9.81 t2 = 2590

t= 27.133

[tex]x=vt=27.133 \times 37.3 \neq 24700 m[/tex]

what else is wrong? :rolleyes:
 
  • #11
rl.bhat
Homework Helper
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Here plane moves with uniform velocity in the vertical and horizontal direction because Σay = g.and Σax is also zero.
And
y component of the velocity:
viy=> (135/3.6)m/s cos 5.9 = 37.3
is not correct. It should be
vy = v*sinθ.
So t = h/v*sinθ
and distance d = t*v*cosθ.
 
  • #12
1,266
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Here plane moves with uniform velocity in the vertical and horizontal direction because Σay = g.and Σax is also zero.
And
y component of the velocity:
viy=> (135/3.6)m/s cos 5.9 = 37.3
is not correct. It should be
vy = v*sinθ.
So t = h/v*sinθ
and distance d = t*v*cosθ.
Thank you so much, it worked out perfectly. :)

I have another relevant question (my last question): Find the magnitude of the rate with which the loaded plane is losing gravitational potential energy.

Here I know that I must use the equation [tex]U_g=mgy[/tex] and to find the rate, I guess I need to find the rate at which y is changing. I tried to divide the height by time taken but that didn't work:

135 km/h × 3.6 = 486 m/s

[tex]t = \frac{h}{v sin \theta} = \frac{2590}{486 sin (5.9)} = 51.8 s[/tex]

[tex]\Delta y = \frac{2590}{51.8} = 50[/tex]

[tex]U_g=(1280 \times 9.81) 50 = 627840[/tex]

But here the answer isn't right (correct answer = 49.1). I think there is something wrong with the way I calculated the change in y, isn't it?
 
  • #13
rl.bhat
Homework Helper
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135 km/h × 3.6 = 486 m/s
Check this calculation.
 
  • #14
1,266
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135 km/h × 3.6 = 486 m/s
Check this calculation.
Why? It seems correct (according to my calculator). I was converting from to kilometers per hour to meters per second, because this is what I did for the previous parts of the question.
 
  • #15
rl.bhat
Homework Helper
4,433
5
Why? It seems correct (according to my calculator). I was converting from to kilometers per hour to meters per second, because this is what I did for the previous parts of the question.
No. It should be
(km/h)/3.6 = m/s
 
  • #16
1,266
11
right...

135 km/h ÷ 3.6 = 37.5 m/s

[tex]t = \frac{2590}{37.5 sin (5.9)} = 671.9 s[/tex]

[tex]\Delta y = \frac{2590}{671.9} = 3.85[/tex]

Ug=mgy= (1280)(9.81)(3.85) = 48343.68

Still, not the right answer. I think there might be something wrong with my value for [tex]\Delta y[/tex] (the rate of change of y), isn't it? :confused:
 
  • #17
1
0
The answer you calculated is in W
if you convert it into kW (the unit required for this question) you will have the correct answer :)
 

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