/*! 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 36 A small rock is thrown verticall... [FREE SOLUTION] | 91Ó°ÊÓ

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A small rock is thrown vertically upward with a speed of 18.0 \(\mathrm{m} / \mathrm{s}\) from the edge of the roof of a 30.0 -m-tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected. (a) What is the speed of the rock just before it hits the street? (b) How much time elapses from when the rock is thrown until it hits the street?

Short Answer

Expert verified
(a) The speed just before hitting the street is 28.0 m/s. (b) The total time in air is 4.90 seconds.

Step by step solution

01

Write the Given Information

We are given that the initial velocity of the rock is \( v_0 = 18.0 \, \text{m/s} \), the height of the building is \( h = 30.0 \, \text{m} \), and we want to find the final velocity just before the rock hits the street and the total time in the air. We will use the acceleration due to gravity as \( g = 9.81 \, \text{m/s}^2 \).
02

Determine the Final Velocity Using Energy Conservation

We can use the conservation of energy to find the speed just before impact. The total mechanical energy at the initial point equals the total mechanical energy just before the rock hits the ground.\[\text{Kinetic Energy at start} + \text{Potential Energy at start} = \text{Kinetic Energy at end}\]This simplifies to:\[ \frac{1}{2}m v_0^2 + mgh = \frac{1}{2}m v^2\]Cancelling the mass \( m \), we have: \[ \frac{1}{2}(18.0)^2 + 9.81 \cdot 30.0 = \frac{1}{2}v^2\]Solving for \( v \), you get \( v = 27.98 \, \text{m/s} \approx 28.0 \, \text{m/s} \).
03

Calculate the Time to Reach Maximum Height

Use the kinematic equation for time to reach maximum height where final velocity at maximum height is 0:\[ v_t = v_0 - gt = 0 \]\[ t = \frac{v_0}{g} = \frac{18.0}{9.81} \approx 1.83 \, \text{s} \]
04

Determine Maximum Height from Roof

Use the kinematic equation to find the additional height reached:\[ h_{max} = \frac{v_0^2}{2g} = \frac{(18.0)^2}{2 \times 9.81} \approx 16.5 \, \text{m} \]
05

Calculate Total Drop from Maximum Height

Add this maximum height to the building height to get the total height from which the rock falls:\[ \text{Total Height} = 30.0 + 16.5 = 46.5 \, \text{m} \]
06

Calculate Time to Fall from Maximum Height

Use the kinematic equation for free fall: \[ h = \frac{1}{2}gt^2 \]\[ 46.5 = \frac{1}{2} \times 9.81 \times t_{fall}^2 \]Solving for \( t_{fall} \), you find \( t_{fall} \approx 3.07 \, \text{s} \).
07

Calculate Total Time in Air

Combine the time to reach maximum height and time to fall back:\[ \text{Total Time} = t_{up} + t_{fall} = 1.83 + 3.07 = 4.90 \, \text{s} \]

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that describes the motion of objects without considering the forces causing the motion. In the case of the rock being thrown upward, we can understand its trajectory through equations that relate velocity, acceleration, and time.
The initial velocity (\(v_0 = 18.0 \text{ m/s}\)) is the speed at which the rock is thrown. As it moves upward, gravity affects its motion by slowing it down until it stops for an instant at its peak height. Here, the final velocity momentarily becomes zero.
Using the kinematic equation \(v_t = v_0 - gt\), we determine how long it takes for the rock to reach this peak. This is crucial for understanding how long it stays airborne. After reaching the peak, gravity accelerates the rock downward, increasing its speed until it impacts the ground, leaving us with an initial upward trajectory followed by a downward descent.
So, the study of kinematics provides a means to predict how long the rock remains in the air, and its speed in various phases of its motion.
Energy Conservation
Energy conservation is a fundamental principle used to solve problems involving projectile motion, especially when no external work is done on the system, like in our rock-throwing scenario. This principle states that the total mechanical energy remains constant if only conservative forces, like gravity, are at play.
Initially, the rock possesses both kinetic energy \(\frac{1}{2}m v_0^2\) due to its motion and potential energy \(mgh\) due to its position above the ground. As the rock ascends, its kinetic energy decreases while potential energy increases until it reaches maximum height. At this point, potential energy is at its peak, and kinetic energy is minimal.When the rock descends, the energy conversion reverses, potential energy converts back into kinetic energy, thus increasing the speed of the rock as it falls. By applying the equation \(\frac{1}{2}m v_0^2 + mgh = \frac{1}{2}m v^2\), and solving for the final velocity, we find the speed of the rock just before hitting the ground is approximately 28.0 m/s. This illustrates how energy conservation helps quantify changes in velocity during projectile motion.
Free Fall
Free fall describes the motion of an object under the influence of gravity alone, with air resistance negligible. Once the rock reaches the peak of its flight, it falls under the exclusive influence of gravity. This phase is crucial as it determines how fast and how far the rock travels before hitting the ground.
During free fall, the rock accelerates downwards at \(g = 9.81 \text{ m/s}^2\), gaining speed as it moves. To calculate the time it takes to fall from its peak, we use the kinematic formula \(h = \frac{1}{2}gt^2\), where \(h = 46.5 \text{ m}\) is the total fall height. Solving for \(t\) gives us the time from the peak to the street.By understanding free fall dynamics, we determine not only how long the rock remains in the air but also how its speed and trajectory are affected by height and gravitational force. This comprehensive insight into free fall adds clarity to the predicted landing time and impact speed of any projectile moving under gravity.

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Most popular questions from this chapter

You throw a glob of putty straight up toward the ceiling, which is 3.60 \(\mathrm{m}\) above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.50 \(\mathrm{m} / \mathrm{s}\). (a) What is the speed of the putty just before it strikes the ceiling? (b) How much time from when it leaves your hand does it take the putty to reach the ceiling?

A world-class sprinter accelerates to his maximum speed in 4.0 s. He then maintains this speed for the remainder of a \(100-\mathrm{m}\) race, finishing with a total time of 9.1 \(\mathrm{s}\) . (a) What is the runner's average acceleration during the first 4.0 \(\mathrm{s} ?\) (b) What is his average acceleration during the last 5.1 \(\mathrm{s} ?\) (c) What is his average acceleration for the entire race? (d) Explain why your answer to part (c) is not the average of the answers to parts (a) and (b).

A car travels in the \(+x\) -direction on a straight and level road. For the first 4.00 s of its motion, the average velocity of the car is \(v_{\mathrm{av}-x}=6.25 \mathrm{m} / \mathrm{s} .\) How far does the car travel in 4.00 \(\mathrm{s}\) ?

Earthquake Analysis. Earthquakes produce several types of shock waves. The most well known are the P-waves (P for primary or pressure) and the S-waves (S for secondary or shear). In the earth's crust, the P-waves travel at around \(6.5 \mathrm{km} / \mathrm{s},\) while the S-waves move at about 3.5 \(\mathrm{km} / \mathrm{s} .\) The actual speeds vary depending on the type of material they are going through. The time delay between the arrival of these two waves at a seismic recording station tells gologists how far away the earthquake occurred. If the time delay is 33 s, how far from the seismic station did the earthquake occur?

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.20 \(\mathrm{m} / \mathrm{s}^{2} .\) At the same same instant a truck, traveling with a constant speed of \(20.0 \mathrm{m} / \mathrm{s},\) overtakes and passes the car. (a) How far beyond its starting point does the car overtake the truck? (b) How fast is the car traveling when it overtakes the truck? (c) Sketch an \(x-t\) graph of the motion of both vehicles. Take \(x=0\) at the intersection. (d) Sketch a \(v_{x}-t\) graph of the motion of both vehicles.

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