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il63 [147K]
2 years ago
14

A plane travels down a runway 2750 m before it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1

.8 km since its wheels left the ground. What is the displacement of the plane?
Physics
1 answer:
ss7ja [257]2 years ago
5 0

Answer: 4.236km

Explanation:

Let's define the point (x, y) as:

x = horizontal distance moved.

y = vertical distance moved.

If the plane starts in the point (0, 0) then:

"A plane travels down a runway 2750 m before it lifts off..."

At this time, the position will be:

P = (0 + 2750m, 0) = (2750m, 0).

"it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1.8 km since its wheels left the ground."

In this case, as the angle is measured from the horizontal, the components will be:

x = 1.8km*cos(37°) = 1.438km

y = 1.8km*sin(37°) =  1.083 km

Then the new position is:

P = (2750m + 1.438 km, 0 + 1.083 km)

Let's write it using the same units for all the quantities:

we know that

1km = 1000m

Then:

2750m = (2750/1000) km = 2.750 km.

Then we can write the new position as:

P = (2.750 km + 1.438km, 1.083km) = (4.188km, 1.083km)

Now, we define the displacement as the distance between the final position and the initial position.

The distance between two points (a, b) and (c, d) is:

D = √( (a  c)^2 + (b - d)^2)

In this case the points are:

(0, 0) for the initial position

(4.188km, 1.083km) for the final position.

And the displacement will be:

D = √( (4.188km - 0)^2 + (1.083 - 0)^2) = 4.236km

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Answer:

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                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

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- Using conservation of energy for initial release point and point where sphere leaves cylinder:

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                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

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