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Chapter 4: Problem 21

A dart is thrown horizontally with an initial speed of \(10 \mathrm{~m} / \mathrm{s}\) toward point \(P\), the bull's-eye on a dart board. It hits at point \(Q\) on the rim, vertically below \(P, 0.19 \mathrm{~s}\) later. (a) What is the distance \(P Q ?\) (b) How far away from the dart board is the dart released?

Short Answer

Expert verified
(a) \(PQ = 0.177 \, \text{m}\). (b) The dart is released 1.9 meters away from the dartboard.

Step by step solution

01

Analyze Horizontal Motion

The dart is thrown horizontally with an initial speed of \(v_{ix} = 10 \, \text{m/s}\). Since there is no horizontal acceleration, the horizontal distance \(d\) traveled by the dart in time \(t = 0.19 \, \text{s}\) is given by \(d = v_{ix} \times t\).
02

Calculate Horizontal Distance

Using the formula for horizontal distance, \(d = v_{ix} \times t\), we substitute \(v_{ix} = 10 \), and \(t = 0.19\):\[d = 10 \times 0.19 = 1.9 \, \text{m}.\]The dart is released 1.9 meters away from the dartboard.
03

Analyze Vertical Motion

The dart falls freely under the influence of gravity. In this case, the initial vertical velocity \(v_{iy} = 0\) (since it's thrown horizontally), the time \(t = 0.19 \, \text{s}\), and acceleration due to gravity \(a = 9.8 \, \text{m/s}^2\). The vertical distance \(P Q\) can be found using the formula \(y = v_{iy}t + \frac{1}{2}a t^2\).
04

Calculate Vertical Distance

Substituting \(v_{iy} = 0\), \(a = 9.8\), and \(t = 0.19\) into the formula \(y = v_{iy}t + \frac{1}{2}a t^2\), we have:\[ y = 0 \times 0.19 + \frac{1}{2} \times 9.8 \times (0.19)^2 = \frac{1}{2} \times 9.8 \times 0.0361 \approx 0.177 \, \text{m}.\] The distance \(P Q\) is 0.177 meters.

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Key Concepts

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

Horizontal Motion
Horizontal motion describes the path an object travels parallel to the ground. In our projectile motion exercise, the dart was thrown horizontally with a speed of 10 meters per second. The fact that it moves horizontally is crucial because it means there's no acceleration affecting its horizontal speed. This is an example of uniform motion.
When talking about horizontal motion, key formulas come into play. A primary formula is the distance calculation:
  • Horizontal Distance, \(d\), is calculated by \(d = v_{ix} \times t\) where \(v_{ix}\) is the initial horizontal velocity, and \(t\) is the time in seconds.
Applying this to the example, with an initial speed of 10 m/s and a time spent in motion of 0.19 seconds, the dart travels 1.9 meters horizontally before hitting the board.
Vertical Motion
Vertical motion deals with how objects move vertically under the influence of gravity. Unlike horizontal motion, vertical motion is affected by gravitational acceleration. The dart, when thrown, starts with no initial vertical speed but soon falls downward due to gravity.
For vertical motion, we often use the following formula:
  • Vertical Distance, \(y\), can be determined using \(y = v_{iy}t + \frac{1}{2}a t^2\) where \(v_{iy}\) is the initial vertical velocity (usually 0 in many horizontal throws), \(a\) is the acceleration due to gravity, and \(t\) is time.
In the case of the dart, it falls 0.177 meters vertically. Understanding how vertical motion works with gravity helps us see why objects thrown horizontally eventually hit the ground.
Initial Velocity
Initial velocity is the speed at which an object starts its journey. It represents both the magnitude and direction of an object's motion. In the horizontal throw of the dart, the initial velocity is 10 meters per second, purely horizontal.
With initial velocity:
  • Vertical Component: Often starts at 0 for horizontal projections.
  • Horizontal Component: Maintains constant speed in ideal conditions (no air resistance).
Using initial velocity, especially in horizontal motion, helps predict how far an object will travel. In physics, decomposing motion into horizontal and vertical components forms the basis for understanding two-dimensional motion.
Acceleration Due to Gravity
Gravity constantly affects objects in vertical motion by applying a downward force. This is denoted by the acceleration due to gravity, usually approximated as 9.8 meters per second squared on Earth's surface. It acts uniformly on all objects, causing them to accelerate downwards.
In our dart example:
  • It causes the dart to fall from the horizontal path as soon as it's released.
  • Determines how fast an object speeds up as it descends.
Knowing the acceleration due to gravity allows accurate calculations of the vertical distance an object travels over time. It played a crucial role in determining the 0.177 meters that the dart fell in the problem. Understanding gravity's impact is essential for mastering projectile motion concepts.

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Most popular questions from this chapter

A cameraman on a pickup truck is traveling westward at 20 \(\mathrm{km} / \mathrm{h}\) while he records a cheetah that is moving westward \(30 \mathrm{~km} / \mathrm{h}\) faster than the truck. Suddenly, the cheetah stops, turns, and then runs at \(45 \mathrm{~km} / \mathrm{h}\) eastward, as measured by a suddenly nervous crew member who stands alongside the cheetah's path. The change in the animal's velocity takes \(2.0 \mathrm{~s}\). What are the (a) magnitude and (b) direction of the animal's acceleration according to the cameraman and the (c) magnitude and (d) direction according to the nervous crew member?

A ball is shot from the ground into the air. At a height of \(9.1 \mathrm{~m}\), its velocity is \(\vec{v}=(7.6 \hat{\mathrm{i}}+6.1 \mathrm{j}) \mathrm{m} / \mathrm{s}\), with \(\hat{\mathrm{i}}\) horizontal and upward. (a) To what maximum height does the ball rise? (b) What total horizontal distance does the ball travel? What are the (c) magnitude and (d) angle (below the horizontal) of the ball's velocity just before it hits the ground?

The acceleration of a particle moving only on a horizontal \(x y\) plane is given by \(\vec{a}=3 t \hat{\mathrm{i}}+4 t \hat{\mathrm{j}}\), where \(\vec{a}\) is in meters per secondsquared and \(t\) is in seconds. At \(t=0\), the position vector \(\vec{r}=(20.0 \mathrm{~m}) \hat{\mathrm{i}}+(40.0 \mathrm{~m}) \hat{\mathrm{j}}\) locates the particle, which then has the velocity vector \(\vec{v}=(5.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(2.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}} \cdot\) At \(t=4.00 \mathrm{~s}\), what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the positive direction of the \(x\) axis? 1

A particle starts from the origin at \(t=0\) with a velocity of \(8.0 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}\) and moves in the \(x y\) plane with constant acceleration \((4.0 \hat{1}+2.0 \hat{j}) \mathrm{m} / \mathrm{s}^{2}\). When the particle's \(x\) coordinate is \(29 \mathrm{~m}\), what are its (a) \(y\) coordinate and (b) speed?

A rugby player runs with the ball directly toward his opponent's goal, along the positive direction of an \(x\) axis. He can legally pass the ball to a teammate as long as the ball's velocity relative to the field does not have a positive \(x\) component. Suppose the player runs at speed \(4.0 \mathrm{~m} / \mathrm{s}\) relative to the field while he passes the ball with velocity \(\vec{v}_{B P}\) relative to himself. If \(\vec{v}_{B P}\) has magnitude \(6.0 \mathrm{~m} / \mathrm{s}\), what is the smallest angle it can have for the pass to be legal?
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