Skip to main content
Ch 03: Motion in Two or Three Dimensions
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 03: Motion in Two or Three DimensionsProblem 19b
Chapter 3, Problem 19b

In a carnival booth, you can 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 m from this point (Fig. E3.19). If you toss the coin with a velocity of 6.4 m/s at an angle of 60° above the horizontal, the coin will land in the dish. Ignore air resistance. What is the vertical component of the velocity of the quarter just before it lands in the dish?
A person tossing a quarter at 60° to hit a dish 2.1 m away, aiming to win a stuffed giraffe.

Verified step by step guidance
1
First, resolve the initial velocity of the quarter into its horizontal and vertical components. The horizontal component \( v_{x} \) can be found using \( v_{x} = v \cdot \cos(\theta) \), where \( v = 6.4 \text{ m/s} \) and \( \theta = 60^{\circ} \). The vertical component \( v_{y} \) is given by \( v_{y} = v \cdot \sin(\theta) \).
Next, calculate the time \( t \) it takes for the quarter to travel the horizontal distance to the dish. Use the formula \( t = \frac{d}{v_{x}} \), where \( d = 2.1 \text{ m} \) is the horizontal distance.
Now, determine the vertical component of the velocity just before the quarter lands in the dish. Use the kinematic equation \( v_{y_{f}} = v_{y} - g \cdot t \), where \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity.
Consider the direction of the vertical component of the velocity. Since the quarter is moving upwards initially and then downwards, the final vertical velocity will be directed downwards.
Finally, ensure that the calculated vertical component of the velocity is consistent with the direction of motion just before landing in the dish, which should be downward.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
9m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Projectile Motion

Projectile motion refers to the motion of an object that is launched into the air and is subject to the force of gravity. It can be analyzed in two dimensions: horizontal and vertical. The horizontal motion is uniform, while the vertical motion is influenced by gravity, resulting in a parabolic trajectory. Understanding this concept is crucial for determining the path and final velocity of the quarter as it travels to the dish.
Recommended video:
Guided course
04:44
Introduction to Projectile Motion

Components of Velocity

Velocity can be broken down into horizontal and vertical components, especially when dealing with angled launches. For a projectile launched at an angle, the initial velocity can be split into two parts: the horizontal component (v_x) and the vertical component (v_y). These components can be calculated using trigonometric functions, where v_x = v * cos(θ) and v_y = v * sin(θ). This breakdown is essential for analyzing the quarter's motion in both dimensions.
Recommended video:
Guided course
05:53
Calculating Velocity Components

Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration, such as gravity. These equations relate displacement, initial velocity, final velocity, acceleration, and time. In this scenario, the vertical motion of the quarter can be analyzed using these equations to find the vertical component of the velocity just before it lands in the dish. The relevant equation for vertical motion is v_y^2 = v_{y0}^2 + 2a_y d_y, where a_y is the acceleration due to gravity.
Recommended video:
Guided course
08:25
Kinematics Equations
Related Practice
Textbook Question

In a carnival booth, you can 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 m from this point (Fig. E3.19). If you toss the coin with a velocity of 6.4 m/s at an angle of 60° above the horizontal, the coin will land in the dish. Ignore air resistance. What is the height of the shelf above the point where the quarter leaves your hand?

3
views
Textbook Question

A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0° above the horizontal. The shot hits the ground 2.08 s later. Ignore air resistance. How far did she throw the shot horizontally?

1
views
Textbook Question

A 124 kg balloon carrying a 22 kg basket is descending with a constant downward velocity of 20.0 m/s. A 1.0 kg stone is thrown from the basket with an initial velocity of 15.0 m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. That person sees the stone hit the ground 5.00 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 m/s. How high is the balloon when the rock is thrown?

4
views
Textbook Question

A 124 kg balloon carrying a 22 kg basket is descending with a constant downward velocity of 20.0 m/s. A 1.0 kg stone is thrown from the basket with an initial velocity of 15.0 m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. That person sees the stone hit the ground 5.00 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 m/s. At the instant the rock hits the ground, how far is it from the basket?

6
views
Textbook Question

A man stands on the roof of a 15.0-m-tall building and throws a rock with a speed of 30.0 m/s at an angle of 33.0° above the horizontal. Ignore air resistance. Calculate Draw x-t, y-t, vx–t, and vy–t graphs for the motion.

2
views
Textbook Question

A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0° above the horizontal. The shot hits the ground 2.08 s later. Ignore air resistance. What are the components of the shot's velocity at the beginning and at the end of its trajectory?

1
views