A volleyball is spiked so that it has an initial velocity of 16.6 m/s directed downward at an angle of 55.2 ° below the horizontal. What is the horizontal component of the ball's velocity when the opposing player fields the ball?
To find the horizontal component of the ball's velocity when the opposing player fields the ball, we need to first determine the vertical and horizontal components of the initial velocity.
Given:
Initial velocity magnitude (v) = 16.6 m/s
Angle below the horizontal (θ) = 55.2°
The vertical component of the initial velocity (v_y) can be found using the formula:
v_y = v * sin(θ)
Substituting the given values:
v_y = 16.6 m/s * sin(55.2°)
v_y ≈ 16.6 m/s * 0.819
v_y ≈ 13.5894 m/s
The horizontal component of the initial velocity (v_x) can be found using the formula:
v_x = v * cos(θ)
Substituting the given values:
v_x = 16.6 m/s * cos(55.2°)
v_x ≈ 16.6 m/s * 0.574
v_x ≈ 9.5184 m/s
Therefore, the horizontal component of the ball's velocity when the opposing player fields the ball is approximately 9.5184 m/s.
To find the horizontal component of the ball's velocity when the opposing player fields the ball, we need to analyze the given information about the initial velocity.
The initial velocity of the ball can be broken down into its horizontal and vertical components.
Given:
Initial velocity (V₀) = 16.6 m/s
Angle below the horizontal (θ) = 55.2°
The horizontal component of the velocity (Vx) can be found using the equation:
Vx = V₀ * cos(θ)
where V₀ is the initial velocity and θ is the angle below the horizontal.
Now, let's plug in the values and calculate:
Vx = 16.6 m/s * cos(55.2°)
Using a calculator, we find:
Vx = 16.6 m/s * 0.5741
Vx ≈ 9.53 m/s
Therefore, the horizontal component of the ball's velocity when the opposing player fields the ball is approximately 9.53 m/s.