To solve this problem, we can use the kinematic equations of motion under constant acceleration. Since the only force acting on the volleyball after it's thrown is gravity, the acceleration $a$ is equal to $-9.8 \, \text{m/s}^2$.

Let’s define the following:

• $t_1 = 0.300 \, \text{s}$: Time at which the volleyball passes the top of the net.
• $t_2 = 0.700 \, \text{s}$: Time it takes from passing the net to reaching the maximum height.
• Total time from launch to the highest point: $t_{\text{total}} = t_1 + t_2 = 1.000 \, \text{s}$.

We are asked to find the height of the net. Here are the steps:

Step 1: Calculate the initial velocity

At the maximum height, the final velocity $v_{\text{max}} = 0$. Using the equation of motion: $v_f = v_i + a t$ where:

• $v_f = 0 \, \text{m/s}$ (at maximum height),
• $v_i$ is the initial velocity we need to find,
• $a = -9.8 \, \text{m/s}^2$,
• $t_{\text{total}} = 1.000 \, \text{s}$.

Plugging in values: $0 = v_i + (-9.8)(1.000)$ $v_i = 9.8 \, \text{m/s}$

Step 2: Calculate the height of the net

Now that we know the initial velocity, we can use the equation for displacement: $d = v_i t + \frac{1}{2} a t^2$ where:

• $t = 0.300 \, \text{s}$ (time to reach the net),
• $v_i = 9.8 \, \text{m/s}$,
• $a = -9.8 \, \text{m/s}^2$.

Substituting the values: $d = (9.8)(0.300) + \frac{1}{2} (-9.8)(0.300)^2$ $d = 2.94 - 0.441$ $d = 2.499 \, \text{m}$

Thus, the height of the net is approximately 2.5 meters.

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Would you like further details on this or have any questions?

Here are 5 related questions to extend your understanding:

1. How would the height of the net change if the initial velocity were higher?
2. What is the maximum height the volleyball reaches above the net?
3. How would air resistance affect the motion of the volleyball?
4. What is the total time the volleyball stays in the air before returning to the ground?
5. If the net were lower, how would that affect the time it takes to reach it?

Tip: When solving problems involving motion, always clearly define the direction of acceleration and velocities to avoid sign errors.