/*! 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 21 A stone is thrown by aiming dire... [FREE SOLUTION] | 91Ó°ÊÓ

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A stone is thrown by aiming directly at the center \(P\) of a picture hanging on a wall. The stone leaves from the starting point horizontally with a speed of \(6.75 \mathrm{~m} / \mathrm{s}\) and strikes the target at point \(Q\), which is \(5.00 \mathrm{~cm}\) below \(P\). Find the horizontal distance between the starting point of the stone and the target.

Short Answer

Expert verified
The horizontal distance is approximately 0.682 meters.

Step by step solution

01

Analyze the Vertical Motion

Let's first analyze the vertical motion of the stone. Since the stone is thrown horizontally, its initial vertical velocity is zero. By the time it reaches point \(Q\), it's moved \(5.00 \, \text{cm} = 0.05 \, \text{m}\) down.We know the acceleration due to gravity is \(g = 9.81 \, \text{m/s}^2\). To find the time \(t\) it takes to fall this vertical distance, we use the equation for uniformly accelerated motion:\[0.05 = \frac{1}{2} \cdot 9.81 \cdot t^2\]
02

Calculate the Time of Flight

Solve for \(t\) in the equation:\[0.05 = \frac{1}{2} \cdot 9.81 \cdot t^2\]\(0.05 = 4.905t^2\)Thus, \(t^2 = \dfrac{0.05}{4.905}\)Solving for \(t\), we get:\[t = \sqrt{\dfrac{0.05}{4.905}} \approx 0.101 \, \text{s}\]
03

Analyze the Horizontal Motion

Now that we have the time of flight \(t = 0.101 \, \text{s}\), we calculate the horizontal distance. The stone moves horizontally at a constant speed of \(6.75 \, \text{m/s}\). The horizontal distance \(d\) can be found using:\[d = v \cdot t = 6.75 \, \text{m/s} \times 0.101 \, \text{s}\]
04

Calculate the Horizontal Distance

Solving for \(d\), we calculate:\[d = 6.75 \times 0.101 \approx 0.682 \text{ m}\]Thus, the horizontal distance between the stone's starting point and the target is approximately \(0.682 \, \text{m}\).

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Key Concepts

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

Uniformly Accelerated Motion
Uniformly accelerated motion refers to a type of motion where an object's velocity changes at a consistent rate due to a constant force.
In the context of projectile motion, this force is usually gravity.
The main equations involve:
  • Final velocity: \(v = u + at\)
  • Displacement: \(s = ut + \frac{1}{2}at^2\)
  • Final velocity squared: \(v^2 = u^2 + 2as\)
The exercise focuses on vertical motion as uniformly accelerated due to gravity, where initial vertical velocity \(u\) is zero.
The stone falls down due to gravity, and the time to reach the target is calculated using the formula for displacement.
Here, displacement is the vertical drop of the stone, and acceleration is due to gravity.
This basic principle helps us determine how far and fast an object will fall under gravity alone.
Horizontal Velocity
Horizontal velocity in projectile motion is constant.
There are no horizontal forces acting on the stone, assuming air resistance is neglected.
Thus, the stone travels horizontally at a uniform speed.
This contrasts with vertical motion, which is affected by gravity.
In our example, the stone has an initial horizontal velocity of \(6.75 \, \text{m/s}\).
This speed doesn't change while the stone is in the air.
Knowing the horizontal velocity and the time of flight allows us to calculate the horizontal distance.
  • Formula: \(d = v \times t\)
This demonstrates a crucial part of the projectile motion, where the horizontal and vertical components operate independently of each other.
Vertical Displacement
Vertical displacement measures how far an object moves vertically from its starting position.
In the given problem, the vertical displacement is \(5 \, \text{cm}\) or \(0.05 \, \text{m}\), as the stone travels downward to hit point \(Q\).
This displacement results from the influence of gravity only and tells us the vertical path of the stone.
It helps calculate how long the stone is in the air using the uniformly accelerated motion formula.
Vertical displacement provides key information for determining other aspects of the motion, like time and impact velocity.
To further understand projectile motion, recognize that such displacement calculations are essential for evaluating the object's behavior as it moves vertically under gravity.
Acceleration due to Gravity
Acceleration due to gravity, denoted as \(g\), is a fundamental force causing objects to fall toward Earth.
It has a constant value of approximately \(9.81 \, \text{m/s}^2\).
This constant acceleration importance is tied to the uniformly accelerated motion of objects like the stone in this example.
Gravity only affects the vertical component of motion, not the horizontal.
While the stone is in the air, gravity constantly accelerates it downwards, explaining the stone's vertical displacement.
This constant accelerative force is why projectile problems use the same formulas for calculating fall time, part of the predictable uniform motion caused by gravity.
  • Key Formula: \(s = \frac{1}{2}gt^2\)
    • Understanding this allows for accurate predictions of an object's vertical motion, crucial for calculating trajectories in various contexts.

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Most popular questions from this chapter

You throw a ball toward a wall at speed \(25.0 \mathrm{~m} / \mathrm{s}\) and at angle \(\theta_{0}=\) \(40.0^{\circ}\) above the horizontal (Fig. 426). The wall is distance \(d=22.0 \mathrm{~m}\) from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall? (d) When it hits, has it passed the highest point on its trajectory?

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