Question

An object has a horizontal velocity component of 6 m/s and a vertical velocity component of 4 m/s.

Answer 1

For the object that has a horizontal velocity component of 6 m/s and a vertical velocity component of 4 m/s we have:

1. The velocity of the object is 7.21 m/s.

2. The angle it makes with the horizontal is 33.7°.

1. The velocity of the object can be found as follows:

$v = \sqrt{v_{x}^{2} + v_{y}^{2}}$

Where:

$v_{x}$: is the horizontal component of the velocity = 6 m/s

$v_{y}$: is the vertical component of the velocity = 4 m/s

Hence, the velocity is:

$v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(6 m/s)^{2} + (4 m/s)^{2}} = 7.21 m/s$

2. The angle it makes with the horizontal can be calculated with the following trigonometric function:

$tan(\theta) = \frac{v_{y}}{v_{x}}$

Where:

θ: is the angle it makes with the horizontal

Therefore, the angle is:

$\theta = tan^{-1}(\frac{v_{y}}{v_{x}}) = tan^{-1}(\frac{4}{6}) = 33.7$

You can learn more about the components of the velocity here: brainly.com/question/2285233?referrer=searchResults

I hope it helps you!