document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1166451>Question 1166451</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 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. <BR>\n" );
document.write( "Determine the value of (a)θ, (b) Vo, and determine the time to reach the highest point with reference from the horizontal surface.\r<BR>\n" );
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document.write( "https://imgur.com/a/OZg4z8I </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #852754 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2138)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=852754\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 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 ($y_A = 0$).\r<BR>\n" );
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document.write( "## 📐 Assumptions Based on Standard Volleyball Court Dimensions\r<BR>\n" );
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document.write( "* **Net Height ($y_C$):** $2.43 \text{ m}$ (Standard men's net height).<BR>\n" );
document.write( "* **Net Distance ($x_C$):** $9.0 \text{ m}$ (Distance from the service line to the net).<BR>\n" );
document.write( "* **Baseline Distance ($x_B$):** $18.0 \text{ m}$ (Total court length, from service line to the far baseline).<BR>\n" );
document.write( "* **Final Height ($y_B$):** $0 \text{ m}$ (Landing on the far baseline).<BR>\n" );
document.write( "* **Gravity ($g$):** $9.81 \text{ m/s}^2$.\r<BR>\n" );
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document.write( "The trajectory equation for the ball is:<BR>\n" );
document.write( "$$y = (\tan\theta)x - \frac{g x^2}{2 V_0^2 \cos^2\theta}$$\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "## (a) Determine the Value of $\theta$ (Launch Angle)\r<BR>\n" );
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document.write( "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 ($x_C = 9 \text{ m}$).\r<BR>\n" );
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document.write( "By substituting the final landing point $B(x_B, 0)$ into the trajectory equation and simplifying, we get the equation for $V_0^2$:<BR>\n" );
document.write( "$$V_0^2 = \frac{g x_B}{\sin(2\theta)}$$\r<BR>\n" );
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document.write( "Substituting this expression for $V_0^2$ into the trajectory equation for the net $C(x_C, y_C)$, and using the identity $\sin(2\theta) = 2\sin\theta\cos\theta$:\r<BR>\n" );
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document.write( "$$y_C = \tan\theta \left(x_C - \frac{x_C^2}{x_B}\right)$$\r<BR>\n" );
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document.write( "Now, substitute the known dimensions:<BR>\n" );
document.write( "$$2.43 \text{ m} = \tan\theta \left(9.0 \text{ m} - \frac{(9.0 \text{ m})^2}{18.0 \text{ m}}\right)$$<BR>\n" );
document.write( "$$2.43 = \tan\theta (9.0 - 4.5)$$<BR>\n" );
document.write( "$$2.43 = \tan\theta (4.5)$$<BR>\n" );
document.write( "$$\tan\theta = \frac{2.43}{4.5} = 0.54$$\r<BR>\n" );
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document.write( "$$\theta = \arctan(0.54)$$\r<BR>\n" );
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document.write( "$$\mathbf{\theta \approx 28.36^\circ}$$\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "## (b) Determine the Value of $V_0$ (Initial Velocity)\r<BR>\n" );
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document.write( "We use the simplified range equation (derived earlier) and the calculated angle $\theta$:<BR>\n" );
document.write( "$$V_0^2 = \frac{g x_B}{\sin(2\theta)}$$\r<BR>\n" );
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document.write( "1.  Calculate $\sin(2\theta)$:<BR>\n" );
document.write( "    $$\sin(2\theta) = \sin(2 \times 28.36^\circ) = \sin(56.72^\circ) \approx 0.8361$$\r<BR>\n" );
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document.write( "2.  Solve for $V_0$:<BR>\n" );
document.write( "    $$V_0^2 = \frac{9.81 \text{ m/s}^2 \times 18.0 \text{ m}}{0.8361}$$<BR>\n" );
document.write( "    $$V_0^2 \approx 211.206 \text{ m}^2/\text{s}^2$$<BR>\n" );
document.write( "    $$V_0 = \sqrt{211.206}$$\r<BR>\n" );
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document.write( "$$\mathbf{V_0 \approx 14.53 \text{ m/s}}$$\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "## (c) Determine the Time to Reach the Highest Point\r<BR>\n" );
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document.write( "The time to reach the highest point ($t_{peak}$) occurs when the vertical velocity component ($V_y$) is zero.\r<BR>\n" );
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document.write( "$$t_{peak} = \frac{V_{0y}}{g}$$\r<BR>\n" );
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document.write( "1.  Calculate the initial vertical velocity component ($V_{0y}$):<BR>\n" );
document.write( "    $$V_{0y} = V_0 \sin\theta$$<BR>\n" );
document.write( "    $$V_{0y} = 14.533 \text{ m/s} \times \sin(28.36^\circ)$$<BR>\n" );
document.write( "    $$V_{0y} \approx 14.533 \times 0.4750 \approx 6.903 \text{ m/s}$$\r<BR>\n" );
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document.write( "2.  Solve for $t_{peak}$:<BR>\n" );
document.write( "    $$t_{peak} = \frac{6.903 \text{ m/s}}{9.81 \text{ m/s}^2}$$\r<BR>\n" );
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document.write( "$$\mathbf{t_{peak} \approx 0.70 \text{ seconds}}$$ </SPAN>\n" );
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