A volleyball is spiked so that it has an initial velocity of 18 m/s directed downward at an angle of 50¡ã below the horizontal. What is the horizontal component of the ball's velocity when the opposing player fields the ball?
thanks
To find the horizontal component of the velocity, we need to use trigonometry.
Given:
Initial velocity (V) = 18 m/s
Angle (θ) = 50° below the horizontal
The horizontal component of velocity (Vx) can be found using the formula:
Vx = V * cos(θ)
Substituting the given values:
Vx = 18 m/s * cos(50°)
Using a calculator, evaluate:
Vx ≈ 18 m/s * 0.643 = 11.574 m/s
Therefore, the horizontal component of the ball's velocity when the opposing player fields the ball is approximately 11.574 m/s.
To find the horizontal component of the ball's velocity, we need to first determine the magnitude of the initial velocity in the horizontal direction.
Given:
Initial velocity magnitude (v₀) = 18 m/s
Launch angle (θ) = 50° below the horizontal
To find the horizontal component of the velocity (v₀x), we can use the following trigonometric relationship:
v₀x = v₀ * cos(θ)
Let's calculate it!
Step 1: Convert the launch angle to radians
θ (in radians) = θ (in degrees) * π / 180
θ (in radians) = 50° * π / 180
θ (in radians) = 50 * π / 180
θ (in radians) = 0.87 radians (approximately)
Step 2: Calculate the horizontal component of the velocity
v₀x = v₀ * cos(θ)
v₀x = 18 m/s * cos(0.87)
v₀x ≈ 18 m/s * 0.6216
v₀x ≈ 11.1892 m/s
Therefore, the horizontal component of the ball's velocity when the opposing player fields the ball is approximately 11.1892 m/s.