Answered

3 The components of a 15-meters-per-second
velocity at an angle of 60.° above the horizontal
are
(1) 7.5 m/s vertical and 13 m/s horizontal
(2) 13 m/s vertical and 7.5 m/s horizontal
(3) 6.0 m/s vertical and 9.0 m/s horizontal
(4) 9.0 m/s vertical and 6.0 m/s horizontal



Answer :

First draw a picture to help make the problem easier. (see attached)

Next you can use sin and cos to find the x and y components of the velocity.

First lets find the vertical component:
[tex]sin = \frac{opp}{hyp} [/tex]
[tex]sin(60) = \frac{y}{15m/s} [/tex]
Multiply both sides by 15m/s
[tex]15m/s * sin(60) = y[/tex]
y ≈ 13 m/s

Now you can find the horizontal component:
[tex]cos = \frac{adj}{hyp} [/tex]
[tex]cos(60) = \frac{x}{15m/s} [/tex]
Multiply both sides by 15m/s
[tex]15m/s * cos(60) - x[/tex]
x = 7.5

Therefore the answer is 2, the vertical (y) component is 13m/s and the horizontal (x) component is 7.5m/s.
View image SpencerTS
johanrusli

The components of the velocity are :

(2) 13 m/s vertical and 7.5 m/s horizontal

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Further explanation

Vector is a quantity that has a magnitude and direction.

A vector in a cartesian coordinate is represented by an arrow in which the slope of the arrow shows the direction of the vector and the length of the arrow shows the magnitude of the vector.

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A position vector of a point is a vector drawn from the base point of the coordinates O (0,0) to that point.

The addition of two vectors can be done in the following ways:

[tex]\overrightarrow {AB} + \overrightarrow {BC} = \overrightarrow {AC}[/tex]

A negative vector is a vector with the same magnitude but in opposite direction.

[tex]\overrightarrow {AB} = -\overrightarrow {BA}[/tex]

Let's tackle the problem!

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Given:

magnitude of the velocity = v = 15 m/s

direction of the velocity = θ = 60°

Asked:

horizontal component of the velocity = v_x = ?

vertical component of the velocity = v_y = ?

Solution:

Firstly , we will calculate the horizontal component of the velocity as follows:

[tex]v_x = v \cos \theta[/tex]

[tex]v_x = 15 \times \cos 60^o[/tex]

[tex]v_x = 15 \times \frac{1}{2}[/tex]

[tex]\boxed {v_x = 7.5 \texttt{ m/s}}[/tex]

[tex]\texttt{ }[/tex]

Next , we will calculate the vertical component of the velocity as follows:

[tex]v_y = v \sin \theta[/tex]

[tex]v_y = 15 \times \sin 60^o[/tex]

[tex]v_y = 15 \times \frac{1}{2} \sqrt{3}[/tex]

[tex]v_y = 7.5 \sqrt{3} \texttt{ m/s}[/tex]

[tex]\boxed {v_y \approx 13 \texttt{ m/s}}[/tex]

[tex]\texttt{ }[/tex]

Learn more

  • Velocity of Runner : https://brainly.com/question/3813437
  • Kinetic Energy : https://brainly.com/question/692781
  • Acceleration : https://brainly.com/question/2283922
  • The Speed of Car : https://brainly.com/question/568302
  • Magnitude of A Vector : https://brainly.com/question/2678571

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Answer details

Grade: High School

Subject: Physics

Chapter: Vectors

View image johanrusli