# Thread: Angle of Flight

1. ## Angle of Flight

An airplane of mass 1300 kg has an engine failure when flying with an airspeed of 131 km/h at an altitude of 2900 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 131 km/h experiencing a drag force of 1300 N that opposes the direction in which the plane is moving.

Suppose the pilot instead had managed to get the airplane engine started such that he was able to apply full throttle and the airplane climbed along a straight line angled above the horizontal so that it gained altitude at a steady rate of 4.61 m/s. Assuming he was again flying with an airspeed of 131 km/h.

Determine the flight angle above the horizontal the plane is flying.

My attempt: if we draw a triangle, its hypotenuse is the thrust force, its opposite side is the force of gravity. So to find the angle we have to use this

$sin \theta = \frac{mg}{F_{thrust}}$

$\theta = sin^{-1}\frac{12753}{4.07}$

But I can't evaluate $sin^{-1} 3133.4$. The correct answer has to be 7.28°. Can anyone help?

2. Hello demode
Originally Posted by demode
An airplane of mass 1300 kg has an engine failure when flying with an airspeed of 131 km/h at an altitude of 2900 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 131 km/h experiencing a drag force of 1300 N that opposes the direction in which the plane is moving.

Suppose the pilot instead had managed to get the airplane engine started such that he was able to apply full throttle and the airplane climbed along a straight line angled above the horizontal so that it gained altitude at a steady rate of 4.61 m/s. Assuming he was again flying with an airspeed of 131 km/h.

Determine the flight angle above the horizontal the plane is flying.

My attempt: if we draw a triangle, its hypotenuse is the thrust force, its opposite side is the force of gravity. So to find the angle we have to use this

$sin \theta = \frac{mg}{F_{thrust}}$

$\theta = sin^{-1}\frac{12753}{4.07}$

But I can't evaluate $sin^{-1} 3133.4$. The correct answer has to be 7.28°. Can anyone help?
Nearly all of the information in this question is irrelevant. All you need are two facts:

• The airspeed of the aircraft is $131$ km/h
.
• The component of this speed in a vertical direction is $4.61$ m/s

So, expressing this component in km/h, we get:
$\frac{4.61\times3600}{1000}= 131\sin\theta$, where $\theta$ is the angle above the horizontal of the flight path

$\Rightarrow \theta = 7.28^o$

3. Thank you so much! It makes perfect sense. I have another relevant question (my last question):

This is the first part of the question I posted previously:

An airplane of mass 1300 kg has an engine failure when flying with an airspeed of 131 km/h at an altitude of 2900 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 131 km/h experiencing a drag force of 1300 N that opposes the direction in which the plane is moving.
Right after this, we are asked:

Find the magnitude of the rate at which the loaded plane is losing gravitational potential energy.

I know that the gravitational potential energy for this problem is $U=mgy$. The rate of change of $U$ is the rate of change of $mgy$. Since m and g are constants, I just need to find the rate of change of y. Could you show me what method to use for finding the rate of change of y? I really don't get this part...

4. Hello demode
Originally Posted by demode
Thank you so much! It makes perfect sense. I have another relevant question (my last question):

This is the first part of the question I posted previously:
An airplane of mass 1300 kg has an engine failure when flying with an airspeed of 131 km/h at an altitude of 2900 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 131 km/h experiencing a drag force of 1300 N that opposes the direction in which the plane is moving.
Right after this, we are asked:

Find the magnitude of the rate at which the loaded plane is losing gravitational potential energy.

I know that the gravitational potential energy for this problem is $U=mgy$. The rate of change of $U$ is the rate of change of $mgy$. Since m and g are constants, I just need to find the rate of change of y. Could you show me what method to use for finding the rate of change of y? I really don't get this part...
Since you don't know the angle of the downward glide, you can't find the rate of change of $y$. Therefore, you have to look at energy as a whole, and consider the Kinetic Energy as well.

The aircraft is gliding at a constant speed under the action of just two forces: drag and its own weight. There is no change in its KE, and hence the work done by the drag force (which will be negative, since it is in the opposite direction to the motion) is exactly equal to the increase in the gravitational potential energy (which will also be negative, of course, since it is losing height).

We have a drag force of $1300$ N moving at $-131$ km/h ( $=-\frac{131\times1000}{3600}$ m/sec). So the work done by this force is:
$-\frac{131 \times 1000 \times 1300}{3600}=-47306$ Joules per sec
So PE is being lost at the rate of $47306$ J/s; i.e. $47.306$ kW.

5. Originally Posted by Grandad
Hello demodeSince you don't know the angle of the downward glide, you can't find the rate of change of $y$. Therefore, you have to look at energy as a whole, and consider the Kinetic Energy as well.

The aircraft is gliding at a constant speed under the action of just two forces: drag and its own weight. There is no change in its KE, and hence the work done by the drag force (which will be negative, since it is in the opposite direction to the motion) is exactly equal to the increase in the gravitational potential energy (which will also be negative, of course, since it is losing height).

We have a drag force of $1300$ N moving at $-131$ km/h ( $=-\frac{131\times1000}{3600}$ m/sec). So the work done by this force is:
$-\frac{131 \times 1000 \times 1300}{3600}=-47306$ Joules per sec
So PE is being lost at the rate of $47306$ J/s; i.e. $47.306$ kW.

I understood your methd.

But we know the glide angle:

$(mg)sin \theta = drag force$

$\theta = sin^{-1} \frac{1300}{1300 \times 9.81} = 5.85$

So now that we know the glide angle is 5.85°, how do we find the rate of change of y?

6. Hello demode
Originally Posted by demode

I understood your methd.

But we know the glide angle:

$(mg)sin \theta = drag force$

$\theta = sin^{-1} \frac{1300}{1300 \times 9.81} = 5.85$

So now that we know the glide angle is 5.85°, how do we find the rate of change of y?
Having just re-read my previous answer, I realise I didn't word it very well. There are, of course, three forces acting on the aircraft, the third one being the lift force. But this is at right-angles to the direction of motion, and hence does no work. Therefore my answer is correct. PE is being lost at the rate of 47306 J/s.

To get the rate of change of the vertical height $y$, simply divide this rate of change of PE by $mg$. (Can you see why?) So the rate of change of $y$ is
$-\frac{47306}{1300\times9.81}=-3.71$ m/s
The equation you have written down is correct, as is the angle of the flight path. This gives an alternative way of calculating the rate of change of $y$. It is simply the vertical component of the speed, in m/s. Which is:
$-\frac{131 \times 1000}{3600}\sin5.85^o = -3.71$ m/s
I haven't seen the original wording of the question, but if (as you suggest) it asks for the rate of loss of PE and then the rate of change of $y$, I think they expect the first method above to be used. But the second is equally valid.