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A stone with a weight of \(5.29 \mathrm{~N}\) is launched vertically from ground level with an initial speed of \(20.0 \mathrm{~m} / \mathrm{s},\) and the air drag on it is \(0.265 \mathrm{~N}\) throughout the flight. What are (a) the maximum height reached by the stone and (b) its speed just before it hits the ground?

Short Answer

Expert verified
(a) 21.48 meters, (b) 20.0 m/s

Step by step solution

01

Calculate Net Force on the Stone

The stone has weight and an air drag force acting against it. The net force can be calculated as follows: \[ F_{ ext{net}} = W_{ ext{stone}} - F_{ ext{drag}} = 5.29 \, \text{N} - 0.265 \, \text{N} = 5.025 \, \text{N} \]This net force will be used to determine the acceleration experienced by the stone.
02

Determine Mass of the Stone

Using the weight of the stone, we can calculate its mass. The weight is given by \[ W = m imes g \]Rearranging for mass \( m \), where \( g = 9.8 \, \text{m/s}^2 \), gives:\[ m = \frac{W}{g} = \frac{5.29}{9.8} = 0.54 \, \text{kg} \]
03

Calculate Acceleration of the Stone

Now, use the net force to find the acceleration \( a \) using Newton's second law:\[ F_{ ext{net}} = m imes a \]So, \[ a = \frac{F_{ ext{net}}}{m} = \frac{5.025 \, \text{N}}{0.54 \, \text{kg}} = 9.31 \, \text{m/s}^2 \]This is the net acceleration in the upward direction.
04

Find Maximum Height Reached by the Stone

Using the kinematic equation to find the maximum height where final vertical velocity becomes zero:\[ v^2 = u^2 + 2as \]Here, final velocity \( v = 0 \), initial velocity \( u = 20.0 \, \text{m/s} \) and \( s \) is the maximum height. Solving for \( s \):\[ 0 = (20.0)^2 - 2 \times 9.31 \times s \]\[ 400 = 18.62 \times s \]\[ s = \frac{400}{18.62} = 21.48 \, \text{m} \]
05

Calculate the Speed Just Before Stone Hits the Ground

As the stone travels down, use the same kinematic equation considering the entire path from max height to the ground, taking positive downward velocity:\[ v^2 = u^2 + 2as \]where \( u = 0 \), \( a = 9.31 \, \text{m/s}^2 \), and \( s = 21.48 \, \text{m} \):\[ v^2 = 0 + 2 \times 9.31 \times 21.48 \]\[ v^2 = 399 \]\[ v = \sqrt{399} = 19.98 \, \text{m/s} \]
06

Conclusion

The maximum height reached by the stone is approximately 21.48 meters, and its speed just before it hits the ground is approximately 20.0 m/s.

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Key Concepts

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

Kinematics
Understanding kinematics is crucial when analyzing projectile motion. It focuses on describing the motion of objects without considering the forces that cause the motion. For the stone launched vertically, its initial velocity was given as 20.0 m/s. This represents a key parameter in kinematics which involves velocity, acceleration, and displacement.

Projectile motion can be broken down into various phases, such as the upward motion until the maximum height is reached, and the downward motion back to the ground level. During each phase, different kinematic principles apply.

  • Displacement: This is the total vertical movement of the stone. The maximum height attained is part of this displacement.
  • Initial and final velocity: The stone's velocity changes due to the forces acting on it, with initial velocity typically given as a starting point.
  • Time of flight: Although not directly asked in the original problem, understanding kinematics helps estimate how long the stone stays in the air.
Newton's Second Law
Newton's Second Law of motion is a fundamental principle in physics: \( F = ma \). This law states that the force applied on an object is equal to the mass of the object multiplied by its acceleration. Here, it plays a key role in determining how the stone's movement is influenced by its mass and the forces acting upon it.

To solve the problem of the stone's motion, understanding the net force was essential. Given the stone's weight as 5.29 N and air resistance of 0.265 N, the net force was calculated by subtracting the air resistance from the weight.

This net force helped determine the acceleration using the equation \( a = \frac{F_{\text{net}}}{m} \), highlighting how forces influence acceleration directly. Newton's Second Law effectively ties together the mass, acceleration, and net force, providing the basis for calculating subsequent motion properties.
Air Resistance
Air resistance is an opposing force that affects the motion of the stone. Even though the stone is launched with an initial force, air drag subdues some of its velocity. This resistance alters the net force acting in the vertical direction, which ultimately affects acceleration and thereby impacts how high the stone can go.

Air resistance is assumed to be a constant 0.265 N in this scenario. While air resistance typically depends on various factors like velocity and surface area, having a constant value helps simplify the solution in educational problems.

In real-world applications, however, air resistance can change, and understanding this concept is essential for accurately predicting the motion in more complex situations. It can have noticeable effects in scenarios with high-speed motion or large surface areas.
Kinematic Equations
Kinematic equations are applied to solve for various aspects of motion such as the stone's maximum height and speed before impact. These equations provide relationships between velocity, acceleration, displacement, and time. In scenarios like this, these equations become indispensable tools.

When we want to know how high the stone went, the equation \( v^2 = u^2 + 2as \) was used. Here, \( v \) is the final velocity (zero at the peak), \( u \) is the initial velocity (20.0 m/s), \( a \) is the acceleration, and \( s \) is the displacement or height.
To find the velocity just before it hits the ground, the same kinematic principle can be applied in reverse.

  • Upward Phase: Calculate the maximum height with final velocity zero.
  • Downward Phase: Calculate the final speed with a known starting velocity of zero when coming back down.
Employing these equations effectively is crucial for understanding how motion unfolds over time.

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Most popular questions from this chapter

In a circus act, a \(60 \mathrm{~kg}\) clown is shot from a cannon with an initial velocity of \(16 \mathrm{~m} / \mathrm{s}\) at some unknown angle above the horizontal. A short time later the clown lands in a net that is \(3.9 \mathrm{~m}\) vertically above the clown's initial position. Disregard air drag. What is the kinetic energy of the clown as he lands in the net?

To form a pendulum, a \(0.092 \mathrm{~kg}\) ball is attached to one end of a rod of length \(0.62 \mathrm{~m}\) and negligible mass, and the other end of the rod is mounted on a pivot. The rod is rotated until it is straight up, and then it is released from rest so that it swings down around the pivot. When the ball reaches its lowest point, what are (a) its speed and (b) the tension in the rod? Next, the rod is rotated until it is horizontal, and then it is again released from rest. (c) At what angle from the vertical does the tension in the rod equal the weight of the ball? (d) If the mass of the ball is increased, does the answer to (c) increase, decrease, or remain the same?

To make a pendulum, a \(300 \mathrm{~g}\) ball is attached to one end of a string that has a length of \(1.4 \mathrm{~m}\) and negligible mass. (The other end of the string is fixed.) The ball is pulled to one side until the string makes an angle of \(30.0^{\circ}\) with the vertical; then (with the string taut ) the ball is released from rest. Find (a) the speed of the ball when the string makes an angle of \(20.0^{\circ}\) with the vertical and (b) the maximum speed of the ball. (c) What is the angle between the string and the vertical when the speed of the ball is one-third its maximum value?

A skier weighing \(600 \mathrm{~N}\) goes over a frictionless circular hill of radius \(R=20 \mathrm{~m}\) (Fig. 8 -62). Assume that the effects of air resistance on the skier are negligible. As she comes up the hill, her speed is \(8.0 \mathrm{~m} / \mathrm{s}\) at point \(B,\) at angle \(\theta=20^{\circ} .\) (a) What is her speed at the hilltop (point \(A\) ) if she coasts without using her poles? (b) What minimum speed can she have at \(B\) and still coast to the hilltop? (c) Do the answers to these two questions increase, decrease, or remain the same if the skier weighs \(700 \mathrm{~N}\) instead of \(600 \mathrm{~N} ?\)

In Fig. 8 -42, a block of mass \(m=3.20 \mathrm{~kg}\) slides from rest a distance \(d\) down a frictionless incline at angle \(\theta=30.0^{\circ}\) where it runs into a spring of spring constant \(431 \mathrm{~N} / \mathrm{m} .\) When the block momentarily stops, it has compressed the spring by \(21.0 \mathrm{~cm} .\) What are (a) distance \(d\) and (b) the distance between the point of the first block-spring contact and the point where the block's speed is greatest?

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