Question

An airplane is moving at 120 m/s at the angle of 10 degree with x axis, through a 30 m/s cross wind blowing at angle of 260 degree with x axis. determine the resultant velocity of the airplane.

Answer 1

Answer:

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Answer 2

Given:

Velocity of the airplane $v_{a}$= 120 m/s

Direction of the airplane with respect to the x-axis $\[\theta _{a}\]$= $\[10^{\circ}\]$

Velocity of the wind $v_{w}$ = 30 m/s

Direction of the wind with respect to the x-axis $\[\theta _{w}\]$= $\[260^{\circ}\]$

To find:

We have to find the resultant velocity of the airplane.

Solution:

The horizontal and vertical components of a vector V are given as,

$V_{h}$ = $Vcos\theta$

$V_{v}$ = $Vsin \theta$

where,

$V_{h}$ - horizontal component

$V_{v}$ - vertical component

$\theta$ - Angle with x-axis

Also, $V = \sqrt{V_{h} ^{2} +V_{v} ^{2} }$  and  $\theta = tan^{-1} (\frac{V_{v} }{V_{h} } )$

Horizontal component of airplane velocity

$=\[120 \times cos(10^{\circ})\]$

= 118.17 m/s

Horizontal component of wind velocity

$=\[30 \times cos(260^{\circ})\]$

= -4.17 m/s

Sum of horizontal components of velocities

=  118.17 + (-4.17) = 114 m/s

Vertical component of airplane velocity

$=\[120 \times sin(10^{\circ})\]$

= 20.49 m/s

Vertical component of wind velocity

$=\[30 \times sin(260^{\circ})\]$

= -29.54 m/s

Sum of vertical components of velocities

= 20.49 + (-29.54) = -9.05 m/s

Magnitude of Resultant

$=\sqrt{114^{2} +(-9.05)^{2} }$

= 114.36 m/s

Direction of resultant

$=tan^{-1} (\frac{-9.05 }{114} )$

= $\[-4.54^{\circ}\]$ = $\[4.54^{\circ}\]$ (Clockwise to the x-axis)

The resultant velocity of the airplane has a magnitude of 114.36 m/s at an angle of $\[4.54^{\circ}\]$ to the x-axis in a clockwise direction.