A rocket lifts off its launching pas with a vertical velocity of 20. m/s and a horizontal velocity of 60. m/s. What are the magnitude and direction of its resultant velocity?
I drew the x and Y axis. 20 m/s on the Y axis and 30 m/s on the X axis. I know I need to find the angle (North West. I don't know what formula to use.
Xo = 60 m/s.
Yo = 20 m/s.
Vo = ? m/s.
Tan A = Yo/Xo = 20/60 = 0.33333
A = 18.43o = Direction.
Vo = Xo/Cos A = 60/Cos18.43 = 63.2 m/s @ 18.43o
To find the magnitude and direction of the resultant velocity, you can use the Pythagorean theorem and trigonometry.
First, use the Pythagorean theorem to find the magnitude (speed) of the resultant velocity:
Resultant velocity = √(horizontal velocity^2 + vertical velocity^2)
Resultant velocity = √(60^2 + 20^2)
Resultant velocity = √(3600 + 400)
Resultant velocity = √4000
Resultant velocity ≈ 63.25 m/s
Next, use trigonometry to find the direction (angle) of the resultant velocity. Since the vertical velocity is positive (upward), and the horizontal velocity is positive (to the right), the angle will be in the north-west direction.
To find the angle, you can use the inverse tangent function (arctan):
Angle = arctan(vertical velocity / horizontal velocity)
Angle = arctan(20 / 60)
Angle = arctan(1/3)
Using a calculator, you can find the angle to be approximately 18.43 degrees.
Therefore, the magnitude of the resultant velocity is approximately 63.25 m/s, and the direction of the resultant velocity is approximately 18.43 degrees north-west.