Question

20m/s at 275 degrees from the x-axis

Answer 1

So, the magnitude of velocity concerning:

• x-axis approximately 1.74 m/s to the x-positive, and
• y-axis approximately -19.92 m/s (19.92 m/s to the y-negative).

Introduction and Formula Used

Hi! Here I will help you to solve problems related to the velocity vector. In this problem, we know that a particle moves at a certain speed with an angle (
`\sf{\theta}`) on the x-axis. Of course, with conditions like this, velocity can be described on the x or y-axis. The result of the x-axis velocity is horizontal movement. On the other hand, the y-axis velocity is vertical movement. See the equation below to know the magnitude of velocity concerning the x and y-axis.

Velocity respect to the x-axis

`\boxed{\sf{\bold{v_x = v_i \cdot \cos(\theta)}}}`

Velocity respect to the y-axis

`\boxed{\sf{\bold{v_y = v_i \cdot \sin(\theta)}}}`

With the following condition:

• 
  `\sf{v_i}` = the initial velocity
• 
  `\sf{\theta}` = elevation angle
• 
  `\sf{v_x}` = the velocity respect to the x-axis
• 
  `\sf{v_y}` = the velocity respect to the y-axis

Problem Solving

We know that:

• 
  `\sf{v_i}` = the initial velocity = 20 m/s
• 
  `\sf{\theta}` = elevation angle = 225°

What was asked?

• 
  `\sf{v_x}` = the velocity respect to the x-axis = ... m/s.
• 
  `\sf{v_y}` = the velocity respect to the y-axis = ... m/s.

Step by step:

• Find the magnitude of velocity respect to the x-axis

`\sf{v_x = v_i \cdot \cos(\theta)}`

`\sf{v_x = 20 \cdot \cos(275^(\circ))}`

`\sf{\bold{v_x \approx 1.74 \: m/s}}`

• Find the magnitude of velocity respect to the y-axis

`\sf{v_y = v_i \cdot \sin(\theta)}`

`\sf{v_y = 20 \cdot \sin(275^(\circ))}`

`\sf{\bold{v_y \approx -19.92 \: m/s}}`

Conclusion

So, the magnitude of velocity concerning:

• x-axis approximately 1.74 m/s to the x-positive, and
• y-axis approximately -19.92 m/s (19.92 m/s to the y-negative)