Math Problem Statement

A regulation volleyball court is L = 18.0 m long and a regulation volleyball net is d = 2.43 m high. A volleyball player strikes the ball a height h = 1.77 m directly above the back line, and the ball's initial velocity makes an angle theta = 51o with respect to the ground (see the figure). In volleyball, it is often advantageous to serve the ball as hard as possible. If you want the ball to land in the opponent's court, however, there is an upper limit on the initial ball speed for a given contact point. At this maximum speed, the ball just barely makes it over the net and then just barely lands in bounds on the back line of the opponent's court.

Solution

This is a typical problem of projectile motion involving a volleyball serve, with specific conditions. Here's how we can approach this:

Given:

  • Court length: L=18.0mL = 18.0 \, \text{m}
  • Net height: d=2.43md = 2.43 \, \text{m}
  • Initial height of ball: h=1.77mh = 1.77 \, \text{m}
  • Angle of projection: θ=51\theta = 51^\circ
  • The ball must barely clear the net and land on the opponent's backline.

Goal:

We need to determine the maximum initial speed of the ball, v0v_0, that satisfies these conditions.

Step-by-Step Approach:

  1. Projectile motion equations:

    • The horizontal distance covered by the ball is given by: x=v0cos(θ)tx = v_0 \cos(\theta) t where tt is the time of flight, and xx is the horizontal displacement.

    • The vertical position of the ball as a function of time is: y=h+v0sin(θ)t12gt2y = h + v_0 \sin(\theta) t - \frac{1}{2} g t^2 where yy is the vertical displacement, g=9.8m/s2g = 9.8 \, \text{m/s}^2 is the acceleration due to gravity, and h=1.77mh = 1.77 \, \text{m} is the initial height.

  2. Conditions to satisfy:

    • The ball must clear the net at x=9.0mx = 9.0 \, \text{m} (half the court) and y=d=2.43my = d = 2.43 \, \text{m}.
    • The ball must land on the opponent’s backline at x=18.0mx = 18.0 \, \text{m}, with y=0y = 0 (ground level).
  3. Two-step analysis:

    • First, use the condition that the ball must just barely clear the net to find a relationship between the velocity and time.
    • Second, use the condition that the ball must land on the backline to calculate the maximum speed.

Let's calculate the maximum speed v0v_0 by considering these constraints.The solution we obtained for the initial speed v0v_0 is negative, which is not physically meaningful. This suggests a potential issue in the setup of the equations. I'll recheck and correct the approach, making sure to properly handle the time and displacement relationships. Let me proceed with refining the calculation.It seems like I can’t do more advanced data analysis right now. Please try again later. If you have other questions or need help, feel free to ask!

Here are 5 related questions you might find helpful:

  1. What is the general formula for projectile motion?
  2. How does the angle of projection affect the range of a projectile?
  3. What is the significance of gravity in projectile motion problems?
  4. How do we calculate the time of flight for a projectile?
  5. What factors determine whether a volleyball serve is effective?

Tip: When solving projectile motion problems, always break down the motion into horizontal and vertical components. It simplifies complex motion significantly!

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Math Problem Analysis

Mathematical Concepts

Projectile Motion
Kinematics

Formulas

x = v0 * cos(θ) * t
y = h + v0 * sin(θ) * t - (1/2) * g * t^2

Theorems

Equations of motion under gravity
Trigonometric functions in motion

Suitable Grade Level

Grade 11-12 (Physics and Pre-Calculus)