A volleyball is spiked so that it has an initial velocity of 16 m/s directed downward at an angle of 62° below the horizontal. What is the horizontal component of the ball's velocity when the opposing player fields the ball?
______________m/s
To find the horizontal component of the ball's velocity, we need to use trigonometry.
Given:
Initial velocity (v) = 16 m/s
Angle (θ) = 62° below the horizontal
The horizontal component of the velocity (v_x) can be found using the formula:
v_x = v * cos(θ)
Substituting the given values:
v_x = 16 m/s * cos(62°)
Using a calculator, the horizontal component of the velocity would be approximately 7.69 m/s (rounded to two decimal places).
Therefore, the horizontal component of the ball's velocity when the opposing player fields the ball is approximately 7.69 m/s.
To find the horizontal component of the ball's velocity, we need to use the given information about its initial velocity.
The initial velocity of the ball can be split into horizontal and vertical components using trigonometry. The horizontal component is given by:
Horizontal Velocity = Initial Velocity * cos(angle)
Where:
Initial Velocity = 16 m/s (given)
Angle = 62° (given)
Using the given values, we can calculate the horizontal velocity:
Horizontal Velocity = 16 m/s * cos(62°)
Horizontal Velocity ≈ 16 m/s * 0.460
Horizontal Velocity ≈ 7.36 m/s
Therefore, the horizontal component of the ball's velocity when the opposing player fields the ball is approximately 7.36 m/s.