/*! 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 19 Win the Prize. In a carnival boo... [FREE SOLUTION] | 91Ó°ÊÓ

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Win the Prize. In a carnival booth, you win a stuffed giraffe if you toss a quarter into a small dish. The dish is on a shelf above the point where the quarter leaves your hand and is a horizontal distance of 2.1 \(\mathrm{m}\) from this point (Fig. E3.19). If you toss the coin with a velocity of 6.4 \(\mathrm{m} / \mathrm{s}\) at an angle of \(60^{\circ}\) above the horizontal, the coin lands in the dish. You can ignore air resistance. (a) What is the height of the shelf above the point where the quarter leaves your hand? (b) What is the vertical component of the velocity of the quarter just before it lands in the dish?

Short Answer

Expert verified
(a) Height = 1.522 m; (b) Vertical velocity just before landing = -0.891 m/s.

Step by step solution

01

Break down the coin's initial velocity

The coin is thrown with an initial velocity of 6.4 m/s at an angle of 60° above the horizontal. We need to find its horizontal and vertical components of velocity.The horizontal component ( v_{0x} ) is calculated as:\[ v_{0x} = v_0 \cdot \cos(60°) = 6.4 \cdot \frac{1}{2} = 3.2 \, \text{m/s} \]The vertical component ( v_{0y} ) is calculated as:\[ v_{0y} = v_0 \cdot \sin(60°) = 6.4 \cdot \frac{\sqrt{3}}{2} = 5.54 \, \text{m/s} \]
02

Determine time of flight to reach the dish

The distance to the dish is 2.1 m horizontally. Using the horizontal component of velocity, calculate the time ( t ) it takes for the coin to reach this distance:\[ x = v_{0x} \cdot t \]\[ 2.1 = 3.2 \cdot t \]\[ t = \frac{2.1}{3.2} = 0.65625 \, \text{s} \]
03

Calculate the height of the shelf

Using the vertical component of the initial velocity and the time calculated, we can find the vertical displacement (the height of the shelf) using the kinematic equation:\[ y = v_{0y} \cdot t - \frac{1}{2} g t^2 \]where g = 9.8 \, \text{m/s}^2 (acceleration due to gravity).\[ y = 5.54 \cdot 0.65625 - \frac{1}{2} \cdot 9.8 \cdot (0.65625)^2 \]\[ y = 3.636 - 2.114 \approx 1.522 \, \text{m} \]
04

Find the vertical velocity before landing

To find the vertical component of velocity just before landing in the dish, we use:\[ v_y = v_{0y} - g t \]\[ v_y = 5.54 - 9.8 \times 0.65625 \]\[ v_y = 5.54 - 6.43125 = -0.89125 \, \text{m/s} \]The negative sign indicates the direction is downward.

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Key Concepts

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

Kinematics
Kinematics is the branch of mechanics that deals with the motion of objects. It involves analyzing how things move through different dimensions without considering the forces that cause the motion. Key components such as displacement, velocity, and acceleration are explored in kinematics.
In projectile motion problems like the carnival game exercise, kinematics helps break down the motion of the coin into simpler parts. Even though the quarter is moving along a curved path, we can separate this path into horizontal and vertical motions. By using kinematic equations and principles, we can find unknown values like the height of the shelf and the speed of the coin as it nears the dish. These calculations make projectiles one of the fascinating topics in physics due to their real-life application in both simple entertainment and complex engineering scenarios.
Horizontal and Vertical Components
When a projectile like a coin is thrown, its motion can be decomposed into two components: horizontal and vertical. This decomposition is crucial because it simplifies the complex motion into manageable parts.
  • Horizontal Component (\(v_{0x}\)): This is the part of the velocity that affects motion along the horizontal plane. For the coin, we calculate it using the cosine of the angle of projection. In our problem, it was determined as 3.2 m/s.
  • Vertical Component (\(v_{0y}\)): This influences how the coin rises and falls. Calculated with the sine of the angle, it was 5.54 m/s in this scenario. This vertical component changes due to gravity as the coin moves.
Understanding these components allows us to predict the projectile's path and determine parameters such as travel time and maximum height, by addressing the horizontal and vertical motions separately.
Acceleration Due to Gravity
Acceleration due to gravity is a key concept in projectile motion. It is the constant acceleration of an object towards the Earth, and its approximate value is 9.8 m/s².
Gravity primarily affects the vertical motion component. As the coin travels from your hand to the dish, gravity decreases its upward velocity until it reaches the peak of its trajectory. After reaching this peak, the velocity begins to increase in the downward direction. In solving these problems, gravity is used in kinematic equations like \(y = v_{0y} \, t - \frac{1}{2} \, g \, t^2\). This helps calculate the change in vertical position and speed over time. Always remember, the value of gravity is negative in these equations because it acts downward, opposite to the initial direction of the vertical motion.
Angle of Projection
The angle at which an object is projected determines how its initial velocity splits between horizontal and vertical components. In projectile motion, this angle is taken between the direction of the initial velocity and the horizontal plane.
For the carnival game, the angle of projection was 60°. This steep angle provided a higher vertical velocity component compared to the horizontal, leading to a tall, curved trajectory for the coin. Different projection angles adjust the range and height a projectile can achieve. A small angle results in a longer range but lower height, while a larger angle results in a shorter range but higher peak. Using the angle, you can calculate the components of the initial velocity using trigonometric functions like cosine and sine, crucial steps to solving these types of physics problems.

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Most popular questions from this chapter

Two students are canoeing on a river. While heading upstream, they accidentally drop an empty bottle overboard. They then continue paddling for 60 minutes, reaching a point 2.0 \(\mathrm{km}\) farther upstream. At this point they realize that the bottle is missing and, driven by ecological awareness, they turn around and head downstream. They catch up with and retrieve the bottle (which has been moving along with the current) 5.0 \(\mathrm{km}\) downstream from the turn-around point. (a) Assuming a constant paddling effort throughout, how fast is the river flowing? (b) What would the canoe speed in a still lake be for the same paddling effort?

A rocket is fired at an angle from the top of a tower of height \(h_{0}=50.0 \mathrm{m} .\) Because of the design of the engines, its position coordinates are of the form \(x(t)=A+B t^{2}\) and \(y(t)=C+D t^{3},\) where \(A, B, C,\) and \(D\) are constants. Furthermore, the acceleration of the rocket 1.00 s after firing is \(\vec{a}=(4.00 \hat{\imath}+3.00 \hat{J}) \mathrm{m} / \mathrm{s}^{2} .\) Take the origin of coordinates to be at the base of the tower. (a) Find the constants \(A, B, C,\) and \(D,\) including their SI units. (b) At the instant after the rocket is fired, what are its acceleration vector and its velocity? (c) What are the \(x\) -and \(y\) -components of the rocket's velocity 10.0 s after it is fired, and how fast is it moving? (d) What is the position vector of the rocket 10.0 s after it is fired?

Crossing the River I. A river flows due south with a speed of 2.0 \(\mathrm{m} / \mathrm{s} . \mathrm{A}\) man steers a motorboat across the river; his velocity relative to the water is 4.2 \(\mathrm{m} / \mathrm{s}\) due east. The river is 800 \mathrm{m}$ wide. (a) What is his velocity (magnitude and direction) relative to the earth? (b) How much time is required to cross the river? (c) How far south of his starting point will he reach the opposite bank?

A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle (Fig. P3.67). The takeoff ramp was inclined at \(53.0^{\circ},\) the river was 40.0 \(\mathrm{m}\) wide, and the far bank was 15.0 \(\mathrm{m}\) lower than the top of the ramp. The river itself was 100 \(\mathrm{m}\) below the ramp. You can ignore air resistance. (a) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank? (b) If his speed was only half the value found in part (a), where did he land?

A rocket designed to place small payloads into orbit is carried to an altitude of 12.0 \(\mathrm{km}\) above sea level by a converted airliner. When the airliner is flying in a straight line at a constant speed of 850 \(\mathrm{km} / \mathrm{h}\) , the rocket is dropped. After the drop, the airliner maintains the same altitude and speed and continues to fly in a straight line. The rocket falls for a brief time, after which its rocket motor turns on. Once its rocket motor is on, the combined effects of thrust and gravity give the rocket a constant acceleration of magnitude 3.00\(g\) directed at an angle of \(30.0^{\circ}\) above the horizontal. For reasons of safety, the rocket should be at least 1.00 \(\mathrm{km}\) in front of the airliner when it climbs the airliner's altitude. Your job is to determine the minimum time that the rocket must fall before its engine starts. You can ignore air resistance. Your answer should include (i) a diagram showing the flight paths of both the rocket and the airliner, labeled at several points with vectors for their velocities and accelerations; (ii) an \(x-t\) graph showing the motions of both the rocket and the airliner; and (iii) a \(y-t\) graph showing the motions of both the rocket and the airliner. In the diagram and the graphs, indicate when the rocket is dropped, when the rocket motor turns on, and when the rocket climbs through the altitude of the airliner.

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