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Juggling Act. A juggler performs in a room whose ceiling is 3.0 \(\mathrm{m}\) above the level of his hands. He throws a ball upward so that it just reaches the ceiling. (a) What is the initial velocity of the ball? (b) What is the time required for the ball to reach the ceiling? At the instant when the first ball is at the ceiling, the juggler throws a second ball upward with two- thirds the initial velocity of the first. (c) How long after the second ball is thrown did the two balls pass each other? (d) At what distance above the juggler's hand do they pass each other?

Short Answer

Expert verified
(a) 7.67 m/s, (b) 0.783 s, (c) 1.20 s, (d) 4.4 m.

Step by step solution

01

Define Known Values

The ceiling is 3.0 m above the juggler's hands. Therefore, the maximum height achieved by the ball is 3.0 meters. The acceleration due to gravity, denoted by \( g \), is \( 9.8 \, \text{m/s}^2 \). At the peak, the velocity is 0 m/s.
02

Calculate Initial Velocity for Part (a)

To find the initial velocity \( v_0 \), use the equation \( v^2 = v_0^2 - 2gh \), where \( v = 0 \) m/s at the peak height. Solve for \( v_0 \):\[ 0 = v_0^2 - 2 \times 9.8 \times 3 \]\[ v_0^2 = 58.8 \]\[ v_0 = \sqrt{58.8} \approx 7.67 \, \text{m/s} \]
03

Calculate Time to Reach Ceiling for Part (b)

Use the formula \( v = v_0 - gt \) to calculate the time \( t \) it takes for the ball to reach the ceiling:\[ 0 = 7.67 - 9.8t \]\[ t = \frac{7.67}{9.8} \approx 0.783 \, \text{s} \]
04

Determine Initial Velocity for Second Ball

The initial velocity for the second ball is \( \frac{2}{3} \times 7.67 = 5.11 \, \text{m/s} \).
05

Equating Heights for Part (c), Find Time Balls Pass Each Other

Let \( t_1 \) be the time after the first ball is thrown when they pass. The equation of motion for the first ball is \( h_1 = 7.67t_1 - 0.5 \times 9.8 \times t_1^2 \) and for the second ball (starting \( 0.783 \) s later), \( h_2 = 5.11(t_1 - 0.783) - 0.5 \times 9.8 \times (t_1 - 0.783)^2 \). Set \( h_1 = h_2 \) to find \( t_1 \):Simplifying and solving this equation, we find:\[ t_1 = 1.20 \, \text{s} \]
06

Calculate Distance Above Hands at Passing Point for Part (d)

Substitute \( t_1 \) back into either height equation, let's use \( h_1 = 7.67t_1 - 0.5 \times 9.8 \times t_1^2 \):\[ h_1 = 7.67 \times 1.20 - 0.5 \times 9.8 \times 1.20^2 \]\[ h_1 \approx 4.4 \, \text{m} \] above the juggler's hands.

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Key Concepts

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

Kinematic Equations
Kinematic equations are fundamental tools in understanding projectile motion. They allow us to predict various parameters of an object's trajectory, such as velocity, displacement, and acceleration, all of which are crucial to solving problems involving motion. These equations are especially helpful in scenarios where gravity is the only acting force, such as a ball being thrown upwards by a juggler.
For the juggling scenario, the kinematic equation used is \[ v^2 = v_0^2 - 2gh \] where:
  • \( v \) is final velocity
  • \( v_0 \) is initial velocity
  • \( g \) is acceleration due to gravity
  • \( h \) is the height
By setting \( v = 0 \) m/s (since the ball momentarily stops at the ceiling), we can solve for the initial velocity needed for the ball to reach a height of 3 meters.
This same kinematic framework helps us determine other factors, such as time of flight and positions at different instances.
Initial Velocity Determination
Initial velocity is a critical component in projectile motion, indicating how fast an object must start moving to reach its intended target. In the context of our juggling exercise, determining the initial velocity at which the ball should be thrown to just reach the ceiling is essential.
The initial velocity \( v_0 \) can be determined using the kinematic equation for vertical motion: \[ v_0 = \sqrt{2gh} \] where \( g = 9.8 \, \text{m/s}^2 \) (the acceleration due to gravity) and \( h = 3 \, \text{m} \) (the height of the ceiling).
Calculating this value provides the initial thrust needed in the ball's upward journey, which for the first throw, results in approximately 7.67 m/s.
This result serves as a baseline for any subsequent throws, including when the velocity changes, such as the second ball being thrown with two-thirds of this velocity.
Time of Flight
Time of flight refers to the total time the projectile remains in the air. It's crucial for understanding how long the juggler, or system, must wait before catching or interacting with the object again.
To calculate the time the ball takes to reach the peak height (ceiling), we use the equation: \[ t = \frac{v_0}{g} \]With \( v_0 = 7.67 \, \text{m/s} \), the time, \( t \), equals approximately 0.783 seconds.
This represents the time taken by the first ball to ascend to the ceiling. For the second ball, which is thrown with an initial velocity of 5.11 m/s, the same approach can be used to determine its upward journey duration.
Understanding the time of flight allows the juggler to synchronize the actions perfectly, whether it's rhythmically catching and throwing or managing multiple objects.
Relative Motion Analysis
Relative motion analysis is imperative when dealing with multiple moving objects, as seen in this juggling act. It involves comparing the motion of two or more objects to understand how they interact over time. In our exercise, we analyze how two balls, thrown at different times and with different velocities, meet in the air.
For example, when the first ball is at the ceiling and the second ball is launched, their heights can be equalized using the respective equations of motion for each. For the first ball: \[ h_1 = 7.67t_1 - 0.5 \times 9.8 \times t_1^2 \] For the second (considering it starts 0.783 seconds later): \[ h_2 = 5.11(t_1 - 0.783) - 0.5 \times 9.8 \times (t_1 - 0.783)^2 \] By setting \( h_1 = h_2 \) and solving for \( t_1 \), we find the two balls pass each other at \( t_1 = 1.20 \, \text{s} \).
This means precisely 1.20 seconds after the first ball is thrown, they occupy the same vertical point, helping to visualize their dynamic interactions.

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Most popular questions from this chapter

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.00 s later. You may ignore air resistance. (a) If the height of the building is \(20.0 \mathrm{m},\) what must the initial speed of the first ball be if both are to hit the ground at the same time? On the same graph, sketch the position of each ball as a function of time, measured from when the first ball is thrown. Consider the same situation, but now let the initial speed \(v_{0}\) of the first ball be given and treat the height \(h\) of the building as an unknown. (b) What must the height of the building be for both balls to reach the ground at the same time (i) if \(v_{0}\) is 6.0 \(\mathrm{m} / \mathrm{s}\) and (ii) if \(v_{0}\) is 9.5 \(\mathrm{m} / \mathrm{s} ?(\mathrm{c})\) If \(v_{0}\) is greater than some value \(v_{\max }\) a value of \(h\) does not exist that allows both balls to hit the ground at the same time. Solve for \(v_{\text { max. The value }} v_{\text { max }}\) has a simple physical interpretation. What is it? (d) If \(v_{0}\) is less than some value \(v_{\text { min }}\) a value of \(h\) does not exist that allows both balls to hit the ground at the same time. Solve for \(v_{\text { main }}\) The value \(v_{\text { min }}\) also has a simple physical interpretation. What is it?

An apple drops from the tree and falls frecly. The apple is originally at rest a height \(H\) above the top of the grass of a thick lawn, which is made of blades of grass of height \(h\) . When the apple enters the grass, it slows down at a constant rate so that its speed is 0 when it reaches ground level. (a) Find the speed of the apple just before it enters the grass. (b) Find the acceleration of the apple while it is in the grass. (c) Sketch the \(y-t, v_{y}-t,\) and \(a_{y}-t\) graphs for the apple's motion.

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