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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) \(0.176 \, \text{m}\), (b) \(1.9 \, \text{m}\).

Step by step solution

01

Understand the Problem

We are given that a dart is thrown horizontally with an initial speed of \(10 \, \text{m/s}\) and it travels for \(0.19 \, \text{s}\) before hitting the board. We need to find out two things: (a) the vertical distance \(PQ\) and (b) the horizontal distance from the release to the point of impact.
02

Analyze the Vertical Motion

In a horizontal throw, the vertical motion is independent of the horizontal motion and is a result of free fall. Therefore, only gravity affects the vertical movement. The distance fallen is given by the formula: \[ y = rac{1}{2}gt^2 \] where \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)), and \(t\) is the time (\(0.19 \, \text{s}\)).
03

Calculate the Vertical Distance \(PQ\)

Plug the known values into the equation: \[ y = rac{1}{2}\times 9.8\, \text{m/s}^2 \times (0.19 \, \text{s})^2 = 0.17618 \, \text{m} \] So, the distance \(PQ\) is approximately \(0.176 \, \text{m}\).
04

Analyze the Horizontal Motion

The horizontal distance covered is based on the initial speed and the time elapsed, since horizontal velocity is constant because there is no horizontal acceleration. Use the formula: \[ x = vt \] where \(v\) is the horizontal speed (\(10 \, \text{m/s}\)) and \(t\) is time (\(0.19 \, \text{s}\)).
05

Calculate the Horizontal Distance from Dart Release

Plug in the known values:\[ x = 10 \, \text{m/s} \times 0.19 \, \text{s} = 1.9 \, \text{m} \] So, the dart is released approximately \(1.9 \, \text{m}\) away from the dartboard.

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Key Concepts

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

Horizontal Velocity
When an object is thrown or launched horizontally, its horizontal velocity is the speed at which it moves along the horizontal axis. In projectile motion, if there is no air resistance, this horizontal velocity remains constant throughout the motion.
For instance, in our example of the dart, its horizontal velocity is given as 10 meters per second. The dart maintains this speed as it travels from the point of release until it hits the dartboard.
Knowing that the horizontal velocity stays the same is crucial because it allows us to directly calculate how far the dart travels horizontally. This is done using the simple formula:
  • \( x = vt \)
Where \( x \) is the distance traveled horizontally, \( v \) is the horizontal velocity, and \( t \) is the time.By multiplying the horizontal velocity (10 m/s) by the time it takes (0.19 seconds) to reach the dartboard, we find the dart covers a horizontal distance of 1.9 meters.
Vertical Motion
Vertical motion in projectile problems is influenced by gravity and is completely independent of horizontal motion. When an object experiences vertical motion in projectile motion, it is similar to an object dropped straight down, undergoing free fall. The dart in our problem exhibits vertical motion as soon as it is released.
The dart falls from its original height just due to gravity pulling it down. There is no initial vertical velocity since it was thrown horizontally.

Equation for Vertical Motion

The formula to find vertical distance covered due to gravity is:
  • \( y = \frac{1}{2}gt^2 \)
Where \( y \) represents the vertical distance fallen, \( g \) is the acceleration due to gravity (9.8 m/s²), and \( t \) is the time (0.19 seconds). This formula helps us calculate how far down the dart has fallen during its flight, which is approximately 0.176 meters in our scenario.
Gravity
Gravity is an ever-present force that pulls objects toward the Earth, at a constant rate. This force is responsible for the downward acceleration experienced in vertical motion. In the context of projectile motion, gravity acts on objects, making them fall downward as they travel horizontally.
In the exercise we have, the dart is subject to gravity the moment it leaves the thrower’s hand. This force causes the vertical motion, resulting in the dart falling a certain distance from its point of release.

The Role of Gravity in Projectiles

Because of gravity, no matter how fast something moves horizontally, it eventually hits the ground unless another force acts upon it. Gravity provides objects with a constant downward acceleration, which is typically measured as 9.8 meters per second squared on Earth. This forms the basis for calculating vertical distances in projectile motion problems.
Acceleration Due to Gravity
Acceleration due to gravity is the rate at which an object speeds up as it falls. This value is a constant 9.8 meters per second squared for objects on Earth. It means that every second, the speed of a freely falling object will increase by 9.8 meters per second.
In our dart example, the acceleration due to gravity solely affects the dart's vertical component since it's thrown horizontally. The dart's vertical speed increases as it falls, even though it started with no vertical component of velocity.
With a known value of acceleration due to gravity, we can easily use it in various calculations, like the formula for the vertical motion in projectiles:
  • \( y = \frac{1}{2}gt^2 \)
This allows us to accurately predict how far and how fast an object like our dart will fall in a specific time period, given no other forces act other than gravity.
Free Fall
Free fall describes the motion of an object when it is falling solely under the influence of gravity, with no other forces, such as air resistance, acting on it. In our dart problem, even though the dart has a horizontal motion, its vertical motion can be treated as free fall.
This is because, once the dart is released, its vertical component of motion is influenced only by gravity. There’s no starting vertical speed, and there's no air resistance considered in our problem.
A more straightforward way to understand free fall is by imagining dropping a stone from a height. The stone will accelerate downwards at the rate of gravity (9.8 m/s²). Similarly, the dart undergoes a free fall during its flight towards the dartboard. Using the existing free fall principles simplifies solving projectile motion problems by splitting them into horizontal and vertical components.

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Most popular questions from this chapter

In basketball, hang is an illusion in which a player seems to weaken the gravitational acceleration while in midair. The illusion depends much on a skilled player's ability to rapidly shift the ball between hands during the flight, but it might also be supported by the longer horizontal distance the player travels in the upper part of the jump than in the lower part. If a player jumps with an initial speed of \(v_{0}=7.00 \mathrm{~m} / \mathrm{s}\) at an angle of \(\theta_{0}=35.0^{\circ},\) what percent of the jump's range does the player spend in the upper half of the jump (between maximum height and half maximum height \()\) ?

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