We can solve this problem using the principles of kinematics. Let’s break it down step by step.

Known Information:

• Time to reach the top of the net: $t_{\text{net}} = 0.300 \, \text{s}$
• Time to reach the maximum height after passing the net: $t_{\text{max}} = 0.700 \, \text{s}$
• Gravitational acceleration: $g = 9.81 \, \text{m/s}^2$

Step 1: Total time to maximum height

The total time from when the ball was thrown to the maximum height is: $t_{\text{total}} = t_{\text{net}} + t_{\text{max}} = 0.300 \, \text{s} + 0.700 \, \text{s} = 1.000 \, \text{s}$

Step 2: Velocity at the maximum height

At maximum height, the velocity is zero, so we use the equation of motion: $v_{\text{max}} = v_0 - g t_{\text{total}}$ Since $v_{\text{max}} = 0$, we can find the initial velocity $v_0$: $0 = v_0 - g t_{\text{total}}$ $v_0 = g t_{\text{total}} = 9.81 \, \text{m/s}^2 \times 1.000 \, \text{s} = 9.81 \, \text{m/s}$

Step 3: Height of the net

Now we use the kinematic equation to find the height of the net: $h_{\text{net}} = v_0 t_{\text{net}} - \frac{1}{2} g t_{\text{net}}^2$ Substitute the known values: $h_{\text{net}} = (9.81 \, \text{m/s} \times 0.300 \, \text{s}) - \frac{1}{2} (9.81 \, \text{m/s}^2 \times (0.300 \, \text{s})^2)$ $h_{\text{net}} = 2.943 \, \text{m} - \frac{1}{2} (9.81 \, \text{m/s}^2 \times 0.090 \, \text{s}^2)$ $h_{\text{net}} = 2.943 \, \text{m} - 0.44145 \, \text{m}$ $h_{\text{net}} = 2.50155 \, \text{m}$

Thus, the height of the net is approximately $h_{\text{net}} = 2.50 \, \text{m}$.

Would you like further details or explanations?

Here are 5 related questions:

1. How would the solution change if the volleyball were thrown from a higher point than the ground?
2. How does air resistance affect the height reached by the ball?
3. What is the maximum height the volleyball reaches in this scenario?
4. How can you calculate the time for the ball to hit the ground after reaching its maximum height?
5. If the ball was thrown with a different initial velocity, how would that change the height of the net?

Tip: In vertical motion problems, breaking the total motion into parts and using symmetry around the maximum height can simplify calculations.