/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 A juggler throws a bowling pin s... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
The time elapsed until the bowling pin returns to the juggler's hand is 1.67 seconds.

Step by step solution

01

Identify the known quantities

The initial speed \(v_i\) is given as 8.20 m/s. The acceleration \(a\) due to gravity is always -9.8 m/s^2 (it's negative because it's going downward). The final height \(h\) is 0, because the pin returns to the juggler's hands.
02

Choose the correct formula

We will use the equation of motion under gravity: \(v_f = v_i + a*t\). We want to find the time \(t\), thus we can rearrange this formula to: \(t = (v_f - v_i) / a \)
03

Apply the known quantities to the formula

We know that at the highest point of the trajectory, or when the object returns to its starting point, the final velocity \(v_f\) is -8.20 m/s (it's negative because it's going in the downward direction). Therefore, substituting the values into the formula, we get: \(t = (-8.20 - 8.20) / -9.8\).
04

Solve for the unknown quantity

If we do the math, we find that the time \(t\) is 1.67 seconds.

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Key Concepts

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

Initial Speed
When discussing projectile motion in physics, a key factor is the initial speed, which is the speed at which an object starts its journey. For example, in the case of a juggler throwing a bowling pin into the air, the initial speed is crucial because it determines how high the pin will go and how long it will stay in the air before returning to the juggler's hand.

The initial speed, often denoted as \( v_i \) in physics equations, is measured in meters per second (m/s) and is a vector quantity, which means it has both magnitude and direction. In the exercise provided, the initial speed of the bowling pin is \( 8.20 \text{ m/s} \) straight up. This upward initial speed will be counteracted by the force of gravity, which will decelerate the bowling pin until it reaches a temporary stop at the peak of its trajectory before falling back down.
Acceleration Due to Gravity
The acceleration due to gravity is a constant force that affects all objects on Earth. It pulls objects towards the center of the Earth at a constant rate of \( 9.8 \text{ m/s}^2 \), irrespective of their mass. This acceleration is denoted by \( g \) and is always directed downwards toward the Earth's surface. It's important to note that in equations, gravity has a negative sign assigned to it, which represents its downward direction.

Because gravity is a type of acceleration, any object in free fall will experience an increase in velocity in the direction of the gravitational pull - that is, downward. In our problem, the bowling pin thrown upwards will slow down at a rate of \( 9.8 \text{ m/s}^2 \) until it stops momentarily at its highest point, before speeding up again as it falls back towards the juggler's hand.
Equation of Motion
The equation of motion used in the context of projectile motion is a formula that relates an object's initial velocity, final velocity, acceleration, and time. It is a linear relationship showing how an object's velocity changes under constant acceleration. The general form of the equation is \( v_f = v_i + a \times t \) where:
  • \( v_f \) is the final velocity,
  • \( v_i \) is the initial velocity,
  • \( a \) is the acceleration, and
  • \( t \) is the time.

In the case of our juggler, since the pin is thrown straight up and falls back down to the starting point, the final velocity of the pin when it comes back down is the same as the initial velocity, but in the opposite direction (thus with a negative sign). We can rearrange the motion equation to solve for time \( t \) by subtracting \( v_i \) from both sides and then dividing by \( a \), resulting in the new equation \( t = (v_f - v_i) / a \). This formula shows that the time it takes for the pin to return is directly related to both its initial speed and the acceleration due to gravity.

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Most popular questions from this chapter

A tennis ball on Mars, where the acceleration due to gravity is \(0.379 g\) and air resistance is negligible, is hit directly upward and returns to the same level \(8.5 \mathrm{~s}\) later. (a) How high above its original point did the ball go? (b) How fast was it moving just after it was hit? (c) Sketch graphs of the ball's vertical position, vertical velocity, and vertical acceleration as functions of time while it's in the Martian air.

A typical male sprinter can maintain his maximum acceleration for \(2.0 \mathrm{~s}\), and his maximum speed is \(10 \mathrm{~m} / \mathrm{s}\). After he reaches this maximum speed, his acceleration becomes zero, and then he runs at constant speed. Assume that his acceleration is constant during the first \(2.0 \mathrm{~s}\) of the race, that he starts from rest, and that he runs in a straight line. (a) How far has the sprinter run when he reaches his maximum speed? (b) What is the magnitude of his average velocity for a race of these lengths: (i) \(50.0 \mathrm{~m} ;\) (ii) \(100.0 \mathrm{~m}\); (iii) \(200.0 \mathrm{~m} ?\)

In the first stage of a two-stage rocket, the rocket is fired from the launch pad starting from rest but with a constant acceleration of \(3.50 \mathrm{~m} / \mathrm{s}^{2}\) upward. At \(25.0 \mathrm{~s}\) after launch, the second stage fires for \(10.0 \mathrm{~s}\), which boosts the rocket's velocity to \(132.5 \mathrm{~m} / \mathrm{s}\) upward at \(35.0 \mathrm{~s}\) after launch. This firing uses up all of the fuel, however, so after the second stage has finished firing, the only force acting on the rocket is gravity. Ignore air resistance. (a) Find the maximum height that the stage-two rocket reaches above the launch pad. (b) How much time after the end of the stage-two firing will it take for the rocket to fall back to the launch pad? (c) How fast will the stagetwo rocket be moving just as it reaches the launch pad?

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 \(30.0 \mathrm{~m}\) below its starting point \(5.00 \mathrm{~s}\) 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.

A bird is flying due east. Its distance from a tall building is given by \(x(t)=28.0 \mathrm{~m}+(12.4 \mathrm{~m} / \mathrm{s}) t-\left(0.0450 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3} .\) What is the instantaneous velocity of the bird when \(t=8.00 \mathrm{~s} ?\)

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