Question

A rocket is launched northeast with a velocity of 150 meters per second at an angle of 45 degrees above the horizontal. Break down the velocity vector into its x, y, and z components.

What is the horizontal (x) component of the rocket's velocity?

a) 106.07 m/s
b) 75.0 m/s
c) 106.07 m/s (cos 45°)
d) 75.0 m/s (sin 45°)

Answer 1

The horizontal component of the rocket's velocity is 106.07 m/s.

Step-by-step explanation:

To break down the velocity vector of the rocket launched northeast, we need to find its horizontal (x) component. The velocity can be broken down into its x and y components using trigonometry. Since the angle of launch is 45 degrees above the horizontal, the x and y components will be equal. The horizontal component of the rocket's velocity can be calculated using the equation Vx = V * cos(theta), where Vx is the horizontal component, V is the velocity, and theta is the angle of launch. Substituting the given values into the equation, we have Vx = 150 m/s * cos(45°) = 106.07 m/s.