/*! 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 68 A rock is thrown from the roof o... [FREE SOLUTION] | 91Ó°ÊÓ

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A rock is thrown from the roof of a building with a velocity \(v_{0}\) at an angle of \(\alpha_{0}\) from the horizontal. The building has height \(h .\) You can ignore air resistance. Calculate the magnitude of the velocity of the rock just before it strikes the ground, and show that this speed is independent of \(\alpha_{0}\) .

Short Answer

Expert verified
The final speed is \(v_f = \sqrt{v_0^2 + 2gh}\), independent of \(\alpha_0\).

Step by step solution

01

Analyze the Initial Conditions

The initial velocity of the rock is given as \(v_0\) at an angle \(\alpha_0\) from the horizontal. This means the horizontal and vertical components of the initial velocity are \(v_{0x} = v_0 \cos(\alpha_0)\) and \(v_{0y} = v_0 \sin(\alpha_0)\), respectively. The height from which the rock is thrown is \(h\). We'll need to use these components to determine the final velocity of the rock.
02

Determine Final Vertical Velocity

The final vertical velocity can be calculated by using kinematic equations. The vertical motion is influenced by gravity, so the equation \(v^2_{fy} = v^2_{0y} + 2gh\) can be used, where \(g\) is the acceleration due to gravity. This simplifies to \(v^2_{fy} = (v_0 \sin(\alpha_0))^2 + 2gh\).
03

Determine Final Horizontal Velocity

The horizontal motion is not affected by gravity or any other forces in this idealized scenario (ignoring air resistance), thus the horizontal component of the velocity remains unchanged. Therefore, \(v_{fx} = v_{0x} = v_0 \cos(\alpha_0)\).
04

Calculate the Magnitude of the Final Velocity

The magnitude of the velocity as the rock strikes the ground is the vector sum of its horizontal and vertical components: \[v_f = \sqrt{v^2_{fx} + v^2_{fy}}\]. Substituting the expressions from Steps 2 and 3, we get \(v_f = \sqrt{(v_0 \cos(\alpha_0))^2 + ((v_0 \sin(\alpha_0))^2 + 2gh)}\).
05

Simplify to Show Independence from \(\alpha_0\)

Observe that \((v_0 \cos(\alpha_0))^2 + (v_0 \sin(\alpha_0))^2 = v_0^2\). Therefore, the expression for \(v_f\) becomes \(v_f = \sqrt{v_0^2 + 2gh}\). This result shows that the final speed is independent of the angle \(\alpha_0\), confirming the exercise's statement.

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Key Concepts

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

Kinematic Equations
Kinematic equations are a set of formulas that describe the motion of objects, typically in one-dimensional or two-dimensional space. These equations are ideal for calculating the position, velocity, and acceleration of moving objects over time. They are especially useful in the study of projectile motion, where an object is launched into the air and its trajectory is influenced by gravity.

There are several key kinematic equations, but some of the most important ones include:
  • The equation for final velocity: \[ v^2 = u^2 + 2as \]where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the displacement.
  • The equation for displacement in terms of initial velocity and time:\[ s = ut + \frac{1}{2}at^2 \]
  • The equation for velocity after a given time:\[ v = u + at \]
In the projectile motion problem of a rock thrown from a building, these equations are used to separate the motion into horizontal and vertical components. This helps in calculating how the initial velocity influences the rock's path until it hits the ground.
Horizontal and Vertical Components
In projectile motion, the initial velocity of an object can be decomposed into horizontal and vertical components. These components allow for a clear understanding of how the object moves through space separately in two dimensions.

The horizontal component of velocity handles movement parallel to the ground, while the vertical component deals with movement impacted by gravity. At the start, an initial velocity \( v_0 \) at an angle \( \alpha_0 \) results in:
  • Horizontal Velocity: \[ v_{0x} = v_0 \cos(\alpha_0) \] This horizontal component remains constant during the motion since no forces act in this direction (assuming air resistance is negligible).
  • Vertical Velocity: \[ v_{0y} = v_0 \sin(\alpha_0) \] This changes over time due to gravity, which consistently pulls the object downward.
When calculating the rock's trajectory, these components provide essential information to determine its path and final velocity.
Acceleration Due to Gravity
Gravity is a fundamental force that acts on all masses, pulling them toward the center of the Earth. In the context of projectile motion, gravity affects the vertical component of motion exclusively.

The acceleration due to gravity is a constant denoted by \( g \) and typically has a value of approximately \( 9.8 \text{ m/s}^2 \) on Earth's surface. This acceleration is always directed vertically downwards, influencing any object in freefall or projectile motion.
  • It causes the vertical velocity of a projectile to increase in magnitude as it falls downward, or to decrease as it rises.
  • The constant acceleration simplifies calculations using the kinematic equation for vertical velocity: \[ v_{fy} = v_{0y} + gt \]However, the time \( t \) is often derived from other conditions of the motion, like the displacement.
  • When calculating the motion of a thrown rock, we consider how gravity increases its vertical velocity until it crashes to the ground.
The power of gravity in these calculations highlights its unchanging effect, helping us to determine the conditions under which the object reaches its final speed.

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Most popular questions from this chapter

A dog running in an open field has components of velocity \(v_{x}=2.6 \mathrm{m} / \mathrm{s}\) and \(v_{y}=-1.8 \mathrm{m} / \mathrm{s}\) at \(t_{1}=10.0 \mathrm{s}\) . For the time interval from \(t_{1}=10.0\) s to \(t_{2}=20.0\) s, the average acceleration of the dog has magnitude 0.45 \(\mathrm{m} / \mathrm{s}^{2}\) and direction \(31.0^{\circ}\) measured from the \(+x\) -axis toward the \(+y\) -axis. At \(t_{2}=20.0 \mathrm{s}\) (a) what are the \(x\) - and \(y\) -components of the dog's velocity? (b) What are the magnitude and direction of the dog's velocity? (c) Sketch the velocity vectors at \(t_{1}\) and \(t_{2} .\) How do these two vectors differ?

An elevator is moving upward at a constant speed of2.50 \(\mathrm{m} / \mathrm{s} .\) A bolt in the elevator ceiling 3.00 \(\mathrm{m}\) above the elevator floor works loose and falls. (a) How long does it take for the bolt to fall to the elevator floor? What is the speed of the bolt just as it hits the elevator floor (b) according to an observer in the elevator? (c) According to an observer standing on one of the floor landings of the building? (d) According to the observer in part (c), what distance did the bolt travel between the ceiling and the floor of the elevator?

A Ferris wheel with radius 14.0 \(\mathrm{m}\) is turning about a horizontal axis through its center (Fig. E3.29). The linear speed of a passenger on the rim is constant and equal to 7.00 \(\mathrm{m} / \mathrm{s} .\) What are the magnitude and direction of the passenger's acceleration as she passes through (a) the lowest point in her circular motion? (b) The highest point in her circular motion? (c) How much time does it take the Ferris wheel to make one revolution?

A projectile is fired from point \(A\) at an angle above thehorizontal. At its highest point, after having traveled a horizontal distance \(D\) from its launch point, it suddenly explodes into two identical fragments that travel horizontally with equal but opposite velocities as measured relative to the projectile just before it exploded. If one fragment lands back at point \(A,\) how far from \(A\) (in terms of \(D\) ) does the other fragment land?

Canadian geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at speeds up to about 100 \(\mathrm{km} / \mathrm{h}\) . If one such bird is flying at 100 \(\mathrm{km} / \mathrm{h}\) relative to the air, but there is a 40 \(\mathrm{km} / \mathrm{h}\) wind blowing from west to east, (a) at what angle relative to the north-south direction should this bird head so that it will be traveling directly southward relative to the ground? (b) How long will it take the bird to cover a ground distance of 500 \(\mathrm{km}\) from north to south? (Note: Even on cloudy nights, many birds can navigate using the earth's magnetic field to fix the north-south direction.)

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