Question

An entertainer juggles balls while doing other activities. In one act, she throws a ball vertically upward, and while it is in the air, she runs to and from a table 5.50 m away at an average speed of 3.00 m/s, returning just in time to catch the falling ball. (a) With what minimum initial speed must she throw the ball upward to accomplish this feat? (b) How high above its initial position is the ball just as she reaches the table?

Short answer

(a) Initial speed is 18.0 m/s; (b) The ball is 16.52 m high when she reaches the table.

Step-by-step solution

Analyzing the Problem

The entertainer throws a ball and uses the time it takes for the ball to go up and come back down to run to a table and back. We need to find the minimum initial velocity to achieve this.

Calculating Time Taken For Round Trip

The table is 5.50 m away, and she runs at 3.00 m/s. The time for a round trip to the table and back is calculated by the formula: \[ t_{round ext{-}trip} = \frac{2 \times 5.50 \, \text{m}}{3.00 \, \text{m/s}} = 3.67 \, \text{s} \]

Understanding the Motion of the Ball

The total time the ball will be in the air is 3.67 s, where it takes half the time to rise to its peak and the other half to fall back, so \[ t_{up} = \frac{3.67}{2} = 1.835 \, \text{s} \]

Applying Kinematic Equation for Upward Motion

To find the initial speed needed for this time, use the kinematic equation: \[ v = u - gt \] Setting the final velocity \( v = 0 \) at the top of its trajectory and \( g = 9.81 \, \text{m/s}^2 \), we solve \[ 0 = u - 9.81 \times 1.835 \] to find \[ u = 18.0 \, \text{m/s} \]

Finding Maximum Height

To calculate the maximum height reached by the ball, use the equation: \[ v^2 = u^2 - 2g h \] With \( v = 0 \), solve for \( h \): \[ 0 = (18.0)^2 - 2 \times 9.81 \times h \] \[ h = \frac{(18.0)^2}{2 \times 9.81} \approx 16.52 \, \text{m} \]

Considering Height When She Reaches Table

At the halfway mark of the time, the ball is at the top of its trajectory (as the running round trip and ball's up-and-down time are equivalent}. So, when she reaches the table distance, the ball is at its maximum height, calculated as 16.52 m.

Key concepts

Kinematic Equations

The kinematic equations are essential tools in analyzing projectile motion, providing the relationships between different motion variables like velocity, acceleration, and displacement. For vertical motion, these equations simplify the understanding of how an object moves under gravity's influence.

In our juggling scenario, one particular kinematic equation, \(v = u - gt\), was used. Here, \(v\) is the final velocity, \(u\) is the initial velocity, \(g\) is the acceleration due to gravity, and \(t\) is the time. When a ball reaches the peak of its vertical path, the final velocity is zero, making it easier to solve for the initial velocity necessary for the entertainer to complete her trick.

Another handy equation in this setup is \(v^2 = u^2 - 2gh\). This formula allows us to determine the maximum height \(h\) the ball reaches by relating it to the initial velocity and gravitational pull. By rearranging the equation and knowing the velocity at the peak is zero, we can find how high the ball goes. Understanding these equations' roles is key to solving such practical physics problems.

Vertical Motion Analysis

Vertical motion analysis breaks down the projectile movement into an easily understandable form. By isolating the vertical component and examining the object under the sole influence of gravity, we gain clearer insights.

When the entertainer throws the ball straight up, it enters vertical motion where gravity steadily slows the ball until it stops momentarily at the peak. We calculated this unfinished upward journey by setting final velocity \(v = 0\) and determined how much initial speed \(u\) the ball required to achieve this.

The motion can be split into two phases:

• The upward travel, where the ball slows down until it reaches the highest point.
• The downward travel, where after the peak, gravity accelerates the ball back towards the ground.

By focusing solely on vertical motion, one isolates influences of horizontal movement and examines how gravity alone acts on a projectile when air resistance is negligible.

Time of Flight Calculation

The time of flight is a critical factor when analyzing projectile motion, especially in scenarios involving both horizontal and vertical components. In our example, the entertainer must complete a round trip to a table and back in the exact time the ball is in the air.

To calculate this, we look at both the horizontal motion (where the entertainer runs) and vertical motion (where the ball rises and falls). The entire round trip duration, calculated as 3.67 seconds, is our time of flight. This can be split evenly into the ball's ascent and descent time, with each phase taking roughly 1.835 seconds.
Gravitational acceleration \(g = 9.81 \, \text{m/s}^2\) directly influences this vertical flight. The time of flight, therefore, is determined by how much time the ball's initial velocity can counteract gravity before being redirected downward. This strategic division simplifies our calculations and reveals why precise timing is vital in juggling acts.