For women's volleyball the top of the net is 2.24 m above the floor and the court measures 9.0 m by 9.0 m on each side of the net. Using a jump serve, a player strikes the ball at a point that is 3.03 m above the floor and a horizontal distance of 8.4 m from the net.

(a) If the initial velocity of the ball is horizontal, what minimum magnitude must it have if the ball is to clear the net?

(b) What maximum magnitude can it have if the ball is to strike the floor inside the back line on the other side of the net? (Assume a horizontal initial velocity, and that the ball is aimed straight at the net rather than diagonally across the court.)

To answer these questions, we can use the principles of projectile motion. Let's break it down step-by-step:

Step 1: Analyze the motion in the vertical direction.
In this case, the vertical motion of the ball can be treated as a free-fall motion. The initial vertical position is 3.03 m above the floor, and the final vertical position is the height of the net, which is 2.24 m.

Step 2: Determine the vertical displacement.
The vertical displacement can be calculated using the equation:
Δy = y_f - y_i
Δy = 2.24 m - 3.03 m
Δy = -0.79 m

Step 3: Calculate the time of flight.
The time of flight for a projectile can be determined using the equation:
Δy = (1/2)gt^2
where g is the acceleration due to gravity and t is the time of flight.
Since the initial vertical velocity is 0 (ball starts at the highest point), the equation simplifies to:
Δy = (1/2)gt^2
-0.79 m = (1/2)(9.8 m/s^2)t^2
Solving for t, we get:
t^2 = (-0.79 m)(2/9.8 m/s^2)
t^2 = -0.161 sec^2
t = √(-0.161 sec^2) [taking the square root of both sides]
t = ±0.401 sec
Since time cannot be negative, we take t = 0.401 sec.

Step 4: Calculate the horizontal displacement.
The horizontal displacement can be calculated using the equation:
Δx = v_x * t
where v_x is the horizontal component of the velocity and t is the time of flight.
The horizontal component of the velocity is the initial velocity of the ball in the horizontal direction.

(a) What minimum magnitude must the initial velocity have to clear the net?
To clear the net, the horizontal displacement (Δx) should be greater than the width of the court, which is 9.0 m.
Δx = v_x * t
9.0 m = v_x * 0.401 sec
v_x = 9.0 m / 0.401 sec
v_x = 22.44 m/s

The minimum magnitude of the initial velocity in the horizontal direction for the ball to clear the net is 22.44 m/s.

(b) What maximum magnitude can the initial velocity have to strike the floor inside the back line?
To strike the floor inside the back line, the horizontal displacement (Δx) should be less than the length of the court, which is 9.0 m on each side.
Δx = v_x * t
9.0 m = v_x * 0.401 sec
v_x = 9.0 m / 0.401 sec
v_x = 22.44 m/s

The maximum magnitude of the initial velocity in the horizontal direction for the ball to strike the floor inside the back line is also 22.44 m/s.

(a) To find the minimum magnitude of the initial velocity that would allow the ball to clear the net, we can use projectile motion equations.

First, let's find the time it takes for the ball to reach the net horizontally. We can use the horizontal distance and the horizontal component of the initial velocity to calculate the time.

Horizontal distance traveled (d) = 8.4 m
Horizontal component of initial velocity (Vx) = ?
Time taken (t) = ?

The formula for horizontal distance is: d = Vx * t
Solving for t, we get: t = d / Vx

Now, let's consider the vertical motion of the ball. The vertical distance covered by the ball should be greater than the height of the net to clear it.

Vertical distance covered (h) = 2.24 m
Vertical component of initial velocity (Vy) = ?
Time taken (t) = ?
Acceleration due to gravity (g) = 9.8 m/s^2

Using the formula for vertical distance covered, h = Vy * t - (1/2) * g * t^2, and substituting the value of time (t) from the horizontal motion equation, we get:

2.24 = Vy * (d / Vx) - (1/2) * g * (d/Vx)^2

Now, we can solve this equation to find the minimum magnitude of the initial velocity (V) needed to clear the net.

(b) To find the maximum magnitude of the initial velocity that would allow the ball to strike the floor inside the back line on the other side of the net, we need to consider the vertical motion of the ball.

The vertical distance covered (h) is the difference between the height of the ball and the height of the net.

Vertical distance covered (h) = 9.0 m - 2.24 m = 6.76 m

Using the formula for vertical distance covered, h = Vy * t - (1/2) * g * t^2, we can solve this equation to find the maximum magnitude of the initial velocity (V) needed to strike the floor inside the back line on the other side of the net.

a) V= 19.92 m/s

b) V= = 13.31 m/s