Question 1166451: The volleyball player serves the ball from point A with an initial velocity at th angle to the horizontal. If the ball just clear the net C and land on the base line B.
Determine the value of (a)θ, (b) Vo, and determine the time to reach the highest point with reference from the horizontal surface.
https://imgur.com/a/OZg4z8I
Answer by CPhill(2138)   (Show Source):
You can put this solution on YOUR website! This is a complex projectile motion problem that uses the geometric constraints of the volleyball court to solve for the initial parameters. Since the diagram is unavailable, I will use the **standard dimensions of an indoor volleyball court** and assume the serve is initiated from the baseline ($y_A = 0$).
## 📐 Assumptions Based on Standard Volleyball Court Dimensions
* **Net Height ($y_C$):** $2.43 \text{ m}$ (Standard men's net height).
* **Net Distance ($x_C$):** $9.0 \text{ m}$ (Distance from the service line to the net).
* **Baseline Distance ($x_B$):** $18.0 \text{ m}$ (Total court length, from service line to the far baseline).
* **Final Height ($y_B$):** $0 \text{ m}$ (Landing on the far baseline).
* **Gravity ($g$):** $9.81 \text{ m/s}^2$.
The trajectory equation for the ball is:
$$y = (\tan\theta)x - \frac{g x^2}{2 V_0^2 \cos^2\theta}$$
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## (a) Determine the Value of $\theta$ (Launch Angle)
We can use the geometric property that the peak height of the parabolic trajectory occurs exactly halfway between the launch and landing points ($x=0$ and $x=18 \text{ m}$). The net is also exactly at this halfway point ($x_C = 9 \text{ m}$).
By substituting the final landing point $B(x_B, 0)$ into the trajectory equation and simplifying, we get the equation for $V_0^2$:
$$V_0^2 = \frac{g x_B}{\sin(2\theta)}$$
Substituting this expression for $V_0^2$ into the trajectory equation for the net $C(x_C, y_C)$, and using the identity $\sin(2\theta) = 2\sin\theta\cos\theta$:
$$y_C = \tan\theta \left(x_C - \frac{x_C^2}{x_B}\right)$$
Now, substitute the known dimensions:
$$2.43 \text{ m} = \tan\theta \left(9.0 \text{ m} - \frac{(9.0 \text{ m})^2}{18.0 \text{ m}}\right)$$
$$2.43 = \tan\theta (9.0 - 4.5)$$
$$2.43 = \tan\theta (4.5)$$
$$\tan\theta = \frac{2.43}{4.5} = 0.54$$
$$\theta = \arctan(0.54)$$
$$\mathbf{\theta \approx 28.36^\circ}$$
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## (b) Determine the Value of $V_0$ (Initial Velocity)
We use the simplified range equation (derived earlier) and the calculated angle $\theta$:
$$V_0^2 = \frac{g x_B}{\sin(2\theta)}$$
1. Calculate $\sin(2\theta)$:
$$\sin(2\theta) = \sin(2 \times 28.36^\circ) = \sin(56.72^\circ) \approx 0.8361$$
2. Solve for $V_0$:
$$V_0^2 = \frac{9.81 \text{ m/s}^2 \times 18.0 \text{ m}}{0.8361}$$
$$V_0^2 \approx 211.206 \text{ m}^2/\text{s}^2$$
$$V_0 = \sqrt{211.206}$$
$$\mathbf{V_0 \approx 14.53 \text{ m/s}}$$
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## (c) Determine the Time to Reach the Highest Point
The time to reach the highest point ($t_{peak}$) occurs when the vertical velocity component ($V_y$) is zero.
$$t_{peak} = \frac{V_{0y}}{g}$$
1. Calculate the initial vertical velocity component ($V_{0y}$):
$$V_{0y} = V_0 \sin\theta$$
$$V_{0y} = 14.533 \text{ m/s} \times \sin(28.36^\circ)$$
$$V_{0y} \approx 14.533 \times 0.4750 \approx 6.903 \text{ m/s}$$
2. Solve for $t_{peak}$:
$$t_{peak} = \frac{6.903 \text{ m/s}}{9.81 \text{ m/s}^2}$$
$$\mathbf{t_{peak} \approx 0.70 \text{ seconds}}$$
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