A player serves a volleyball striking the volleyball at a 60 degree angle with a velocity of 15 m/s.What is the horizontal velocity of the volleyball 0.5 seconds before it achieves max height? m/sWhat is the magnitude of the vertical velocity the instant right before it contacts an opponent at the same height it was served from? m/sIf the top of the net is 2.25 meters, how high above the net did the ball travel? mHow long was the ball in the air? sHow far did the ball travel horizontally when it was in the air? m

The problem involves breaking down the initial velocity into its horizontal and vertical components, and then using the equations of motion to find the required quantities.

1. The horizontal velocity of the volleyball 0.5 seconds before it achieves max height: The horizontal velocity remains constant throughout the flight and is given by Vx = V * cos(θ), where V is the initial velocity and θ is the angle of projection. So, Vx = 15 * cos(60) = 7.5 m/s.

2. The magnitude of the vertical velocity the instant right before it contacts an opponent at the same height it was served from: The vertical velocity at the maximum height is zero and it increases linearly with time as the ball falls down. Since the ball lands at the same height it was served from, the magnitude of the vertical velocity just before it lands is equal to the initial vertical velocity, which is Vy = V * sin(θ) = 15 * sin(60) = 13 m/s.

3. How high above the net did the ball travel? The maximum height reached by the ball is given by H = Vy^2 / (2g), where g is the acceleration due to gravity. So, H = (13^2) / (29.8) = 8.6 m. Since the height of the net is 2.25 m, the ball travelled 8.6 - 2.25 = 6.35 m above the net.

4. How long was the ball in the air? The total time of flight is given by T = 2Vy / g = 213 / 9.8 = 2.65 s.

5. How far did the ball travel horizontally when it was in the air? The horizontal distance travelled by the ball is given by D = Vx * T = 7.5 * 2.65 = 19.88 m.