Solve Physics Problem: Urgent Cessna-182 Plane Mass 1290kg Glide Angle

[sid88]

Homework Statement

A fully loaded Cessna-182 airplane of mass 1290kg has an engine failure when flying with an airspeed of 139 km/h at an altitude of 2710 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 139km/h experiencing a drag force of 1270 N that opposes the directin in which the plane is moving.
i) Find the magnitude of:

a) The glide angle
5.75 degrees

b) The lift force.
12654.9 N

c) The rate at which the loaded plane is losing gravitational potential energy[/B]

ii) 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 5.61 m/s. Assuming he was again flying with an airspeed of 139km/h determine:

i) the flight angle above the horizontal the plane is flying.

The Attempt at a Solution

I have already found the glide angle and lift force using F = mg cos(theta)
The glide angle is 5.76 degrees
The lift force is 12654.9 N

 

[alphysicist]

Hi sid88,

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, can you find the rate of change of y?

Once you find that, a quick multiplication will give you the rate of change of mgy.

 

[Moetasim]

To find the rate at which the plane is losing gravitational potential energy, we can use the formula P = mgh. We know the mass of the plane (1290kg) and the height (2710m), so we can plug in these values to find the rate of energy loss. This would give us a value of approximately 3.5 x 10^7 J/s.

For part ii), we can use the formula v = u + at to find the velocity of the plane as it climbs. We know the initial velocity (139 km/h) and the acceleration (5.61 m/s^2), so we can plug these values in and solve for the final velocity. This would give us a value of approximately 139.28 km/h.

To find the flight angle above the horizontal, we can use trigonometry and the fact that the plane is traveling at a constant speed of 139 km/h. We can set up a right triangle with the horizontal distance traveled being 139 km and the vertical distance being 5.61 m. Using the tangent function, we can solve for the angle and find that the plane is flying at an angle of approximately 0.024 degrees above the horizontal. This angle is very small, meaning the plane is flying almost parallel to the ground.