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A stone is dropped into a river from a bridge \(43.9 \mathrm{~m}\) above the water. Another stone is thrown vertically down \(1.00 \mathrm{~s}\) after the first is dropped. Both stones strike the water at the same time. (a) What is the initial speed of the second stone?

Short Answer

Expert verified
The initial speed of the second stone is approximately 12.3 m/s.

Step by step solution

01

- Calculate the time it takes for the first stone to fall

Use the equation of motion for free fall: \[ h = \frac{1}{2}gt^2 \] where \( h = 43.9 \, \mathrm{m} \) and \( g = 9.81 \, \mathrm{m/s^2} \). Solve for \( t \). \[ 43.9 = \frac{1}{2} \times 9.81 \times t^2 \] \[ t^2 = \frac{43.9 \times 2}{9.81} \] \[ t^2 = 8.95 \] \[ t = \frac{\text{√}8.95}{1} \] \[ t \thickapprox 2.99 \, \text{s} \]
02

- Calculate the fall time for the second stone

Since the second stone is thrown 1.00 s after the first one is dropped, the time it takes to hit the water is: \( t_{\text{second}} = t - 1.00 \, \text{s} \) \[ t_{\text{second}} = 2.99 \, \text{s} - 1.00 \, \text{s} \] \[ t_{\text{second}} = 1.99 \, \text{s} \]
03

- Determine the initial speed of the second stone

Use the equation of motion with initial velocity for the second stone: \[ h = v_{\text{initial}} t + \frac{1}{2} g t^2 \] Rearrange to solve for \( v_{\text{initial}} \). \[ 43.9 = v_{\text{initial}} \times 1.99 + \frac{1}{2} \times 9.81 \times (1.99)^2 \] \[ 43.9 = v_{\text{initial}} \times 1.99 + \frac{1}{2} \times 9.81 \times 3.9601 \] \[ 43.9 = v_{\text{initial}} \times 1.99 + 19.42 \] \[ 24.48 = v_{\text{initial}} \times 1.99 \] \[ v_{\text{initial}} = \frac{24.48}{1.99} \] \[ v_{\text{initial}} \thickapprox 12.3 \, \text{m/s} \]

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Key Concepts

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

Free Fall Motion
Free fall motion describes the movement of objects under the sole influence of gravity. No other forces, like air resistance, are taken into account. In free fall, all objects accelerate at the same rate regardless of their mass.

This is essential when analyzing how objects fall from a certain height. For instance, if you drop a stone from a bridge, it begins its free fall at zero initial velocity. The only force affecting it is gravity, which pulls it towards the Earth.

As the stone falls, it accelerates uniformly. This is expressed with the gravitational acceleration value, usually denoted as \( g \). In most physics problems near the Earth’s surface, \( g = 9.81 \) meters per second squared (m/s²).

Understanding free fall motion is crucial in solving problems involving objects dropped or thrown vertically.
Equations of Motion
The equations of motion are mathematical formulas that describe the behavior of moving objects. These come in very handy for solving free fall problems and other motion scenarios.

One commonly used equation in free fall motion is \[ h = \frac{1}{2}gt^2 \], where \( h \) is height, \( g \) is gravitational acceleration, and \( t \) is the time taken to fall.

Another useful one when initial speed is involved is \[ h = v_{\text{initial}} t + \frac{1}{2}gt^2 \].

These equations help you determine different variables such as height, time, initial velocity, and final velocity. Familiarity with these formulas enhances your ability to tackle various physics problems involving motion.
Initial Velocity Calculation
Determining the initial velocity of an object in motion can be essential for solving physics problems. Initial velocity (\(v_{\text{initial}}\)) is the speed at which an object starts moving.

In our exercise, you need to find the initial speed of a stone that is thrown down so it hits the water at the same time as another stone in free fall. The relevant equation is \[ h = v_{\text{initial}} t + \frac{1}{2} g t^2 \]. By rearranging this formula, you can solve for \( v_{\text{initial}} \).

Understanding how to calculate the initial velocity helps in predicting and analyzing the future position and speed of moving objects. This is critical not only for academic exercises but also in real-world applications like engineering and various technology fields.
Gravitational Acceleration
Gravitational acceleration is the rate at which an object accelerates due to Earth's gravity. This value is approximately 9.81 m/s² near the Earth's surface.

In free fall motion, this acceleration is what causes the object to increase its velocity as it falls. When analyzing the motion of falling objects, it's crucial to consider gravitational acceleration to predict how long it will take for them to reach the ground and at what speed.

Gravitational acceleration is central to solving equations of motion. It provides a constant that can be used to find other variables such as time, distance, and initial or final velocity in a falling object's trajectory.

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Most popular questions from this chapter

An interstellar ship has a mass of \(1.20 \times 10^{6} \mathrm{~kg}\) and is initially at rest relative to a star system. (a) What constant acceleration is needed to bring the ship up to a speed of \(0.10 c\) (where \(c\) is the speed of light, \(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\) ) relative to the star system in \(3.0\) days? (b) What is that acceleration in \(g\) units? (c) What force is required for the acceleration? (d) If the engines are shut down when \(0.10 c\) is reached (the speed then remains constant), how long does the ship take (start to finish) to journey \(5.0\) light-months, the distance that light travels in \(5.0\) months?

A certain particle has a weight of \(22 \mathrm{~N}\) at a point where \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\). What are its (a) weight and (b) mass at a point where \(g=4.9 \mathrm{~m} / \mathrm{s}^{2} ?\) What are its (c) weight and (d) mass if it is moved to a point in space where \(g=0 ?\)

Then Jumps A \(29.0 \mathrm{~kg}\) child, with a \(4.50 \mathrm{~kg}\) backpack on his back, first stands on a sidewalk and then jumps up into the air. Find the magnitude and direction of the force on the sidewalk from the child when the child is (a) standing still and (b) in the air. Now find the magnitude and direction of the net force on Earth due to the child when the child is (c) standing still and (d) in the air.

An \(85 \mathrm{~kg}\) man lowers himself to the ground from a height of \(10.0 \mathrm{~m}\) by holding onto a rope that runs over a frictionless pulley to a \(65 \mathrm{~kg}\) sandbag. With what speed does the man hit the ground if he started from rest?

Newton's Second Law written in equation form states $$\vec{a}=\frac{\vec{F}^{\text {net }}}{m}$$ Your roommate says "That's silly. Everyone knows it takes a force to keep something moving at a constant velocity, even when there's no acceleration." Do you agree with your roommate? If so, explain why physics classes bother to teach the law. If you disagree, how would you try to convince your roommate of the error of his/her ways?

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