Question

A circus juggler performs an act with balls that he tosses with his right hand and catches with his left hand. Each ball is launched at an angle of \(75^{\circ}\) and reaches a maximum height of \(90 \mathrm{~cm}\) above the launching height. If it takes the juggler \(0.2 \mathrm{~s}\) to catch a ball with his left hand, pass it to his right hand and toss it back into the air, what is the maximum number of balls he can juggle?

Short answer

Short Answer: The maximum number of balls the juggler can juggle is 5.

Step-by-step solution

Calculate the time of flight for a single ball

We can use the kinematic equations for projectile motion to calculate the time of flight. We know the maximum height h and launch angle \(θ = 75°\). We'll start by finding the vertical component of velocity, \(v_y\): Using the equation: $$h = \frac{v^2_y}{2g}$$ Where h is the maximum height (0.9 m), \(v_y\) is the vertical velocity, and g is the acceleration due to gravity (approximately \(9.81 \mathrm{~m/s^2}\)). Solving for \(v_y\), we get: $$v_y = \sqrt{2gh} = \sqrt{(2)(9.81 \mathrm{~m/s^2})(0.9 \mathrm{~m})} ≈ 4.42 \mathrm{~m/s}$$ Now we'll find the total velocity at launch using the angle. We know that: $$v_y = v\sin(θ)$$ So, we can find v by: $$v = \frac{v_y}{\sin(θ)} = \frac{4.42 \mathrm{~m/s}}{\sin(75°)} ≈ 4.55 \mathrm{~m/s}$$ By knowing the total velocity, we can find the horizontal velocity component \(v_x\): $$v_x = v\cos(θ) ≈ (4.55 \mathrm{~m/s})(\cos(75°)) ≈ 1.18 \mathrm{~m/s}$$ Now we have both vertical and horizontal velocity components at launch. To find the time of flight, we'll use the equation: $$t = \frac{2v_y}{g}$$ Solving for t, we get: $$t = \frac{2(4.42 \mathrm{~m/s})}{9.81 \mathrm{~m/s^2}} ≈ 0.9 \mathrm{~s}$$ So, the time of flight for each ball is 0.9 s.

Determine the maximum number of balls that can be juggled

We know it takes 0.2 s for the juggler to catch, pass, and relaunch a ball. So, in the 0.9 s window where a ball is in flight, the juggler needs to have enough time to manage the other balls. Let n be the number of balls. The time taken to catch, pass, and relaunch the n-1 remaining balls must be less than the total time of flight of a single ball: $$(n-1)(0.2 \mathrm{~s}) ≤ 0.9 \mathrm{~s}$$ Solving for n: $$n-1 ≤ \frac{0.9 \mathrm{~s}}{0.2 \mathrm{~s}}$$ $$n-1 ≤ 4.5$$ $$n ≤ 5.5$$ Since the number of balls must be an integer, we can infer that the maximum number of balls the juggler can juggle is 5.

Key concepts

Kinematic Equations

Kinematic equations are essential tools in physics for analyzing motion. These equations allow us to calculate different variables such as time, distance, velocity, and acceleration in both linear and projectile motions. In projectile motion, the motion can be dissected into horizontal and vertical components, which each are analyzed separately using kinematic equations.

• Vertical motion is affected by gravity, while horizontal motion assumes a constant velocity when air resistance is negligible.
• Key kinematic equations include:
  - The equation for displacement: \( d = v_i t + \frac{1}{2} a t^2 \)
  - The equation for final velocity: \( v_f = v_i + a t \)
• For the juggler's problem, we used the vertical motion equation \( h = \frac{v_y^2}{2g} \) to find the vertical component of the initial velocity, and further, \( t = \frac{2v_y}{g} \) to find the time of flight.

These equations helped determine how long each ball stayed in the air and enabled the calculation of other necessary parameters.

Vertical Velocity Component

In projectile motion, the vertical velocity component, denoted as \(v_y\), is critical to understanding how high and how long an object stays in the air. This component is the part of the initial velocity that goes against gravity, determining the maximum height achieved and the time of flight.

• The vertical component is calculated using the equation: \( v_y = v \sin(\theta) \),where \(v\) is the initial velocity and \(\theta\) is the launch angle.
• In our juggler example, we found \(v_y\) using the maximum height equation: \( h = \frac{v_y^2}{2g} \), providing us \(v_y = 4.42 \text{ m/s} \).
• The vertical velocity helps determine the entire flight path, affecting both the ascent and descent, which are symmetrical.

By understanding \(v_y\), the height each ball reaches and how long it stays airborne can be accurately calculated.

Time of Flight

The time of flight in a projectile motion is the total time an object spends traveling through the air from launch to landing. It's dictated primarily by the vertical motion component since gravity influences this duration by pulling the object back down.

• Time of flight depends on the initial vertical velocity and is calculated by the formula: \( t = \frac{2v_y}{g} \), where \(v_y\) is the initial vertical velocity and \(g\) is the acceleration due to gravity.
• For our juggling juggler, knowing that \( t = 0.9 \text{ s} \), tells us each ball is airborne for almost a full second, allowing us to determine how the juggler can manage catching and tossing.
• The time of flight is symmetrical, with half spent ascending and the other half descending.

Understanding time of flight is crucial for timing in juggling, as it directly correlates with how many objects can be in the air simultaneously, impacting the entire act's feasibility.