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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) PQ = 0.1764 m; (b) 1.9 m from the dart board.

Step by step solution

01

Understand the Horizontal Motion

The dart is thrown horizontally at an initial speed of \( v_{0x} = 10 \, \text{m/s} \). Since there is no horizontal acceleration, the horizontal speed remains constant throughout the motion.
02

Calculate the Horizontal Distance

The horizontal distance traveled by the dart can be calculated using the formula \( d_x = v_{0x} \times t \). Here, \( t = 0.19 \, \text{s} \) and \( v_{0x} = 10 \, \text{m/s} \). Substituting these values, \( d_x = 10 \, \text{m/s} \times 0.19 \, \text{s} = 1.9 \, \text{m} \). This is the distance from the release point to the dart board.
03

Understand the Vertical Motion

The dart falls under the influence of gravity, with no initial vertical velocity. The vertical displacement can be determined using the formula \( y = \frac{1}{2} g t^2 \), where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.
04

Calculate the Vertical Distance

Substitute the known values into the vertical displacement equation: \( y = \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (0.19 \, \text{s})^2 \). This simplifies to \( y = 0.1764 \, \text{m} \). This is the vertical distance between points \( P \) and \( Q \).
05

Determine the Distance \( PQ \)

The distance \( PQ \) represents the vertical distance traveled by the dart during its descent, which we calculated earlier to be \( y = 0.1764 \, \text{m} \).

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Key Concepts

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

Horizontal Motion
When dealing with projectile motion, understanding horizontal motion is crucial. Imagine throwing a dart straight forward. This directional movement is known as horizontal motion. A key characteristic of horizontal motion is that the speed remains unchanged throughout the travel. This is because there is no acceleration acting in the horizontal direction.

In the exercise we are analyzing, the dart has an initial horizontal speed of \(10 \, \text{m/s}\). It maintains this speed as it travels towards the dart board. To find out how far the dart travels horizontally, we can use the formula:
\[ d_x = v_{0x} \times t \]
where \(d_x\) is the horizontal distance, \(v_{0x}\) is the initial horizontal speed, and \(t\) is the time of travel. For our dart, the horizontal distance becomes:
  • \(v_{0x} = 10 \, \text{m/s}\) (initial speed)
  • \(t = 0.19 \, \text{s}\) (time taken)
  • \(d_x = 1.9 \, \text{m}\)
This means the dart traveled 1.9 meters before hitting the dart board.
Vertical Motion
Vertical motion, in the context of projectile motion, is how the object moves due to gravity. Starting with no initial vertical velocity, the dart in this scenario gains speed as it falls. This happens due to gravity pulling it downward.

To calculate the vertical displacement, we can use the formula:
\[ y = \frac{1}{2} g t^2 \]
This formula gives us the vertical distance an object falls under gravity. In our exercise, the variables stand for:
  • \(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity)
  • \(t = 0.19 \, \text{s}\) (time in motion)
  • \(y = 0.1764 \, \text{m}\) (vertical distance)
Therefore, Point Q is 0.1764 meters below Point P, defining the precise vertical distance traveled by your dart.
Acceleration Due to Gravity
Acceleration due to gravity is a fundamental concept in projectile motion. It is the constant pull that Earth exerts on objects, dictating how quickly something falls. This acceleration is always directed towards the center of the Earth, acting independently of horizontal motion.

In calculations of motion like our dart problem, gravity's influence is depicted by the constant \(g\), which is \(9.8 \, \text{m/s}^2\). This indicates that every second, the velocity of a falling object increases by 9.8 meters per second in the downward direction.

Even if a dart is thrown horizontally, gravity ensures it will fall vertically as it travels forward.
  • This means the dart will cover a horizontal distance while also descending vertically.
  • Understanding gravity helps to predict how long it takes for a projectile to hit the ground.
  • In our case, this concept allowed us to calculate the vertical fall without needing any initial vertical speed.
By acting on vertical motion, gravity seamlessly brings together the horizontal and vertical components of projectile motion.

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