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:
[tex] v = \sqrt{v_{x}^{2} + v_{y}^{2}} [/tex]
Where:
[tex] v_{x} [/tex]: is the horizontal component of the velocity = 6 m/s
[tex] v_{y}[/tex]: is the vertical component of the velocity = 4 m/s
Hence, the velocity is:
[tex] v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(6 m/s)^{2} + (4 m/s)^{2}} = 7.21 m/s [/tex]
2. The angle it makes with the horizontal can be calculated with the following trigonometric function:
[tex] tan(\theta) = \frac{v_{y}}{v_{x}} [/tex]
Where:
θ: is the angle it makes with the horizontal
Therefore, the angle is:
[tex]\theta = tan^{-1}(\frac{v_{y}}{v_{x}}) = tan^{-1}(\frac{4}{6}) = 33.7[/tex]
You can learn more about the components of the velocity here: https://brainly.com/question/2285233?referrer=searchResults
I hope it helps you!