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Chapter 2: Problem 47

You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, \(6.00 \mathrm{~s}\) after it was thrown. What is the speed of the rock just before it reaches the water \(28.0 \mathrm{~m}\) below the point where the rock left your hand? Ignore air resistance.

Short Answer

Expert verified
The speed of the rock just before it reaches the water is \(23.44 \mathrm{~m/s}\).

Step by step solution

01

Identify Given Variables

In the given problem, the rock falls a vertical distance \(d = 28.0 \mathrm{~m}\) from rest, therefore, the initial velocity \(u = 0 \mathrm{~m/s}\). The acceleration due to gravity \(a = 9.81 \mathrm{~m/s^2}\). We are required to find the final velocity \(v\).
02

Choose the Correct Equation of Motion

Since we know the initial velocity (u), distance (d), and acceleration (a), and we are trying to find the final speed (v), we can use the equation of motion which connects these quantities: \(v^2 = u^2 + 2ad\).
03

Substitute the Given Values

Substitute the given values into the equation: \(v^2 = 0 + 2*9.81*28 \). Therefore, \(v^2 = 549.36 \mathrm{~m^2/s^2}\). Take the square root of both sides to solve for \(v\).
04

Solve for the Final Velocity

Taking the square root of \(549.36 \mathrm{~m^2/s^2}\) gives \(v = 23.44 \mathrm{~m/s}\).
05

Write Down the Final Answer

The speed of the rock just before it hits the water is \(23.44 \mathrm{~m/s}\). Note that this is only the magnitude of the velocity; the direction is downwards, but since we only asked for speed, we ignore the direction.

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Key Concepts

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

Kinematics
Kinematics focuses on the motion without considering the forces causing it. When we analyze the motion of the rock thrown off the bridge, we mainly look at how it travels through space over time without worrying about the forces acting on it.
To simplify, kinematics involves quantities such as:
- Distance: how far an object moves
- Displacement: the change in position of the object
- Velocity: how fast an object moves in a particular direction
- Acceleration: how quickly an object's velocity changes over time.
For our example, we consider the rock鈥檚 journey as it travels vertically, upward and then downward. Its initial velocity when thrown upward, its moment of reaching the highest point, and its subsequent fall to the water are all components of its motion. By understanding these, we can then apply the correct equations to determine how fast it is moving at any given point in its path.
Gravity
Gravity is a crucial force in projectile motion and is mainly responsible for pulling objects downward. In this scenario, the acceleration due to gravity is constant at approximately 9.81 meters per second squared. This gravitational force acts on the rock from the moment it leaves your hand until it hits the water.

When you throw the rock upwards, gravity acts against the upward motion, slowing it down until it stops momentarily at its peak before accelerating downward. It鈥檚 important to note that during its fall, gravity is what causes the rock to speed up as it moves towards the earth.

This consistent gravitational acceleration allows us to predictably calculate how long it takes for the rock to place back into your view and its speed once it reaches the bottom. Understanding gravity is vital in kinematics to accurately predict the motion of objects in freely falling scenarios.
Equations of Motion
Equations of Motion are mathematical formulas that enable us to describe and predict the movement of objects. They connect the core components of kinematics such as displacement, velocity, acceleration, and time. For the task of calculating the final speed of the rock, we used a specific equation:

\[ v^2 = u^2 + 2ad \]

where
  • \(v\) is the final velocity of the rock,
  • \(u\) is the initial velocity, which in this case was provided as zero,
  • \(a\) represents the acceleration due to gravity (9.81 m/s虏),
  • and \(d\) is the vertical distance the rock travels (28 m in this situation).

By substituting these values into the equation, and solving for \(v\), we can determine the rock's speed just before it reaches the water's surface. These equations are vital tools in physics; they enable us to analyze different motion scenarios effectively and find unknown variables where direct measurement is not feasible.

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Most popular questions from this chapter

A rock is thrown straight up with an initial speed of \(24.0 \mathrm{~m} / \mathrm{s}\) Neglect air resistance. (a) At \(t=1.0 \mathrm{~s}\), what are the directions of the velocity and acceleration of the rock? Is the speed of the rock increasing or decreasing? (b) At \(t=3.0 \mathrm{~s}\), what are the directions of the velocity and acceleration of the rock? Is the speed of the rock increasing or decreasing?

Sam heaves a 16 lb shot straight up, giving it a constant upward acceleration from rest of \(35.0 \mathrm{~m} / \mathrm{s}^{2}\) for \(64.0 \mathrm{~cm}\). He releases it \(2.20 \mathrm{~m}\) above the ground. Ignore air resistance. (a) What is the speed of the shot when Sam releases it? (b) How high above the ground does it go? (c) How much time does he have to get out of its way before it returns to the height of the top of his head, \(1.83 \mathrm{~m}\) above the ground?

An object's velocity is measured to be \(v_{x}(t)=\alpha-\beta t^{2}\), where \(\alpha=4.00 \mathrm{~m} / \mathrm{s}\) and \(\beta=2.00 \mathrm{~m} / \mathrm{s}^{3} .\) At \(t=0\) the object is at \(x=0 .\) (a) Calculate the object's position and acceleration as functions of time. (b) What is the object's maximum positive displacement from the origin?

You are on the roof of the physics building, \(46.0 \mathrm{~m}\) above the ground (Fig. \(\mathbf{P 2 . 7 0}\) ). Your physics professor, who is \(1.80 \mathrm{~m}\) tall, is walking alongside the building at a constant speed of \(1.20 \mathrm{~m} / \mathrm{s}\). If you wish to drop an egg on your professor's head, where should the professor be when you release the egg? Assume that the egg is in free fall.

A ball is thrown straight up from the ground with speed \(v_{0}\). At the same instant, a second ball is dropped from rest from a height \(H\) directly above the point where the first ball was thrown upward. There is no air resistance. (a) Find the time at which the two balls collide. (b) Find the value of \(H\) in terms of \(v_{0}\) and \(g\) such that at the instant when the balls collide, the first ball is at the highest point of its motion.
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