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Please Help!!!!
A helicopter flies with an airspeed of 42.5 m/s [W]. If the wind is traveling with a velocity of 25.0 m/s [E30°S] relative to the ground then determine the velocity of the helicopter relative to the ground.

a) Use the data from the question to determine the velocity components for the helicopter’s speed relative to the air and the wind’s speed relative to the ground.

b) Use your velocity components from part (a) and determine the resultant velocity for each component. (In other words, determine the resultant horizontal component for the helicopter’s ground speed and the resultant vertical component for the helicopter’s ground speed).

c) Using your data, determine the resultant velocity of the helicopter relative to the ground.

Respuesta :

Answer:

a) The vertical component of the helicopters speed relative to the air = 0 m/s

The horizontal component of the helicopters speed relative to the air = -42.5 m/s

The vertical component of the wind speed relative to the ground = -21.65 m/s

The horizontal component of the wind  speed relative to the ground = 12.5 m/s

b) The x-component of the helicopter speed relative to the wind speed =  -30 m/s

The y-component of the helicopter speed relative to the wind speed =  -12.5 m/s

c) The resultant speed of the helicopter relative to the ground is 32.5 m/s in the direction of W 22.6° S

Explanation:

The speed of the helicopter = 42.5 m/s

The direction of flight of the helicopter = West

The wind speed = 25.0 m/s

The direction of the wind = E30°S

a) The velocity components of the helicopters speed relative to the air are;

Vertical component = 0 m/s

Horizontal component = -42.5 m/s

The velocity components of the wind speed relative to the ground are;

Vertical component = -25 × cos(30) = -21.65 m/s

Horizontal component = 25 × sin(30) = 12.5 m/s

b) To find the speed of the helicopter relative to the wind we have;

x-component of the helicopter speed = -42.5 m/s

y-component of the helicopter speed = -0

x-component of the wind speed relative to the ground = 25×sin(30) = 12.5 m/s

y-component of the wind speed relative to the ground = -25×cos(30) = -21.65 m/s

Therefore, the x-component of the helicopter speed relative to the wind speed = -42.5 + 12.5= -30 m/s

the y-component of the helicopter speed relative to the wind speed = 0 +  (-12.5) = -12.5 m/s

c) The resultant speed of the helicopter relative to the ground is given by the relation;

[tex]v_{h, ground} =\sqrt{(v_{h,horizontal})^2 + (v_{h,vertical})^2}[/tex]

= √((-30)² + (-12.5)²) =32.5 m/s

The direction of motion from the x-axis is given as follows;

[tex]tan(x) = \dfrac{Vertical \ velocity \ component}{Horizontal \ velocity \ component} = \dfrac{-12.5}{-30} = \dfrac{5}{12}[/tex]

[tex]x = tan^{-1} \left (\dfrac{5}{12} \right ) = 22.62 ^{\circ}[/tex]

Therefore, the helicopter is flying at 32.5 m/s in the direction of W 22.6° S.

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