Question

an airplane is moving at 120 metre per second at an angle of 10 degree with x axis through a 30 metre per second cross Wind blowing at angle of 262 degree with x axis determine the resultant velocity of the airplane​

Answer 1

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Answer 2

Given:

Velocity of the airplane = $120m/s$

Angle of the airplane with X-axis = $10$°

Velocity of wind blowing = $30m/s$

Angle of the blowing wind with X-axis = $262$°

To find:

Resultant velocity of the airplane.

Formula to be used:

• The formula used in this concept can be utilized for various topics like velocities, forces, etc...
• Let the velocity of airplane be denoted by '$A$'.
• Let the velocity of wind be denoted by '$B$'.
• Let the angle between these two velocities be denoted by '$\alpha$'.
• Consider all the angles mentioned above, in anti-clockwise direction from X-axis.
• Now, the resultant of the two above mentioned velocities is be denoted by '$R$' and is given by:

$R=\sqrt{A^{2}+B^{2}+2ABcos\alpha }$......(1)

Note:

An image is attached along with the solution for reference. In image, an angle 'θ' is mentioned, which is considered here as '$\alpha$'. Kindly check with that.

Step-Wise Solution:

Step-1:

As per the given data and indications mentioned above:

• Velocity of the airplane is:  $A=120m/s$
• Velocity of the wind is: $B=30m/s$
• Angle between the two velocities is given by:

$\alpha =262-10$

$\alpha =252$°

⇒ The angles between the two velocities is $252$°, as shown in the image.

Step-2:

• This step involves in the calculation of the resultant velocity.
• Using the equation (1), the resultant velocity is calculated in the following way:

$R=\sqrt{120^{2}+30^{2}+2*120*30*cos252}\\ R=\sqrt{120^{2}+30^{2}+7200(-0.309)}\\ R=\sqrt{14400+900-2224.8}\\ R=\sqrt{13075.2}\\ R=114.35m/s$

Step-3:

• This step involves in the calculation of the angle made by resultant with X-axis.
• Let the angle made by resultant with x-axis be denoted by 'β'. It is calculated using the formula:

$\beta =tan^{-1}((BSin\alpha )/(A+BCos\alpha ))\\\beta =tan^{-1}(30sin252/(120+30cos252))\\\beta =tan^{-1}((-28.53)/110.729)\\\beta =tan^{-1}(-0.258)$

⇒$\beta =-14.45$°

⇒$\beta =360-14.45$

⇒$\beta =345.55$°

Final Answer:

∴ The required resultant velocity of the airplane is $114.35m/s$ and it makes an angle of $345.55$° with X-axis.