A ball is moving at 18ms -1 in a direction inclined at 60° to the horizontal. the horizontal componet of its velocity is?
Vx = 18*Cos60.
To find the horizontal component of the ball's velocity, we need to calculate the value of the velocity vector in the horizontal direction.
Given:
Magnitude of velocity (V) = 18 m/s
Angle of inclination (θ) = 60°
The horizontal component of velocity (Vx) can be found using the formula:
Vx = V * cos(θ)
First, let's convert the angle from degrees to radians:
θ (in rad) = θ (in degrees) * π / 180
θ (in rad) = 60 * π / 180
θ (in rad) = π / 3
Now, we can substitute the values into the formula:
Vx = 18 * cos(π / 3)
Using the value of cos(π / 3) ≈ 0.5:
Vx = 18 * 0.5
Vx = 9 m/s
Therefore, the horizontal component of the ball's velocity is 9 m/s.
To find the horizontal component of the velocity, we need to use trigonometry.
First, let's define the given information:
- The magnitude of the ball's velocity is 18 m/s (18 ms^ -1).
- The angle between the ball's velocity vector and the horizontal direction is 60 degrees.
In a right-angled triangle, the horizontal component of a vector can be found using the cosine of the angle between the vector and the horizontal axis.
So, to calculate the horizontal component, we use the equation:
Horizontal component = Velocity * cos(angle)
Substituting the given values, we have:
Horizontal component = 18 m/s * cos(60°)
Using the cosine of 60 degrees, which is 0.5, we get:
Horizontal component = 18 m/s * 0.5
The calculation simplifies to:
Horizontal component = 9 m/s
Therefore, the horizontal component of the ball's velocity is 9 m/s.