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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

Expert verified
(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

01

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.
02

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: \[ t_{round ext{-}trip} = \frac{2 \times 5.50 \, \text{m}}{3.00 \, \text{m/s}} = 3.67 \, \text{s} \]
03

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 \[ t_{up} = \frac{3.67}{2} = 1.835 \, \text{s} \]
04

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 \[ u = 18.0 \, \text{m/s} \]
05

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 \( h \): \[ 0 = (18.0)^2 - 2 \times 9.81 \times h \] \[ h = \frac{(18.0)^2}{2 \times 9.81} \approx 16.52 \, \text{m} \]
06

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, air bags produce a maximum acceleration of \(60 g\) that lasts for only \(36 \mathrm{~ms}\) (or less). How far (in meters) does a person travel in coming to a complete stop in \(36 \mathrm{~ms}\) at a constant acceleration of \(60 \mathrm{~g} ?\)

A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin returns to the juggler's hand?

An egg is thrown nearly vertically upward from a point near the cornice of a tall building. The egg just misses the cornice on the way down and passes a point 30.0 m below its starting point 5.00 s after it leaves the thrower's hand. Ignore air resistance. (a) What is the initial speed of the egg? (b) How high does it rise above its starting point? (c) What is the magnitude of its velocity at the highest point? (d) What are the magnitude and direction of its acceleration at the highest point? (e) Sketch \(a_y-t, v_y-t\), and \(y-t\) graphs for the motion of the egg.

You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, 6.00 s after it was thrown. What is the speed of the rock just before it reaches the water 28.0 m below the point where the rock left your hand? Ignore air resistance.

You are standing at rest at a bus stop. A bus moving at a constant speed of \(5.00 \mathrm{~m} / \mathrm{~s}\) passes you. When the rear of the bus is \(12.0 \mathrm{~m}\) past you, you realize that it is your bus, so you start to run toward it with a constant acceleration of \(0.960 \mathrm{~m} / \mathrm{~s}^{2}\). How far would you have to run before you catch up with the rear of the bus, and how fast must you be running then? Would an average college student be physically able to accomplish this?

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