An airplane traveling at 200 mph in a direction of 40 degrees north of east encounters a wind blowing east at 50.0 mph. Find the velocity of the plane measured from the ground.
240.5 mph @ E32.3°N
V = 200[40o] + 50[0o].
V = 200*Cos40+200*sin40 + 50
V = 153.2+128.6i + 50 = 203.2 + 128.6i = 240.5mi/h[32.3o] N. of E.
To find the velocity of the plane measured from the ground, we need to consider the vector addition of the velocity of the plane and the velocity of the wind.
Let's break down the given information:
Velocity of the plane (Vp) = 200 mph
Direction of the plane (θp) = 40 degrees north of east
Velocity of the wind (Vw) = 50.0 mph
Direction of the wind (θw) = east
To calculate the velocity of the plane measured from the ground, we need to find the resultant velocity vector by adding the velocities of the plane and the wind.
First, we need to break down the velocities into their horizontal and vertical components.
For the plane:
Velocity of the plane in the x-direction (Vpx) = Vp * cos(θp)
Velocity of the plane in the y-direction (Vpy) = Vp * sin(θp)
For the wind:
Velocity of the wind in the x-direction (Vwx) = Vw * cos(θw)
Velocity of the wind in the y-direction (Vwy) = Vw * sin(θw)
Now, let's calculate the values:
Vpx = 200 mph * cos(40 degrees)
= 200 mph * 0.766
≈ 153.2 mph
Vpy = 200 mph * sin(40 degrees)
= 200 mph * 0.643
≈ 128.6 mph
Vwx = 50.0 mph * cos(0 degrees)
= 50.0 mph * 1.0
= 50.0 mph
Vwy = 50.0 mph * sin(0 degrees)
= 50.0 mph * 0.0
= 0.0 mph
Now, we can add the horizontal and vertical components of the plane and wind velocities to get the resultant velocity measured from the ground.
Resultant velocity in the x-direction (Vrx) = Vpx + Vwx
= 153.2 mph + 50.0 mph
= 203.2 mph
Resultant velocity in the y-direction (Vry) = Vpy + Vwy
= 128.6 mph + 0.0 mph
= 128.6 mph
Finally, we can calculate the magnitude and direction of the resultant velocity using the Pythagorean theorem and trigonometry:
Magnitude of the resultant velocity (Vr) = sqrt(Vrx^2 + Vry^2)
= sqrt((203.2 mph)^2 + (128.6 mph)^2)
≈ 241.5 mph
Direction of the resultant velocity (θr) = arctan(Vry / Vrx)
= arctan(128.6 mph / 203.2 mph)
≈ 33.9 degrees
Therefore, the velocity of the plane measured from the ground is approximately 241.5 mph in a direction of 33.9 degrees north of east.