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Chapter 2: Problem 12

A stone is thrown straight upward and it rises to a maximum height of \(20 \mathrm{~m}\). With what speed was it thrown? Take up as the positive \(y\) -direction. The stone's velocity is zero at the top of its path. Then \(v_{f y}=0, y=20 \mathrm{~m}, a=-9.81 \mathrm{~m} / \mathrm{s}^{2}\). (The minus sign arises because the acceleration due to gravity is always downward and we have taken up to be positive.) Use \(v_{s}^{2}=v_{i}^{2}+2 a y\) to find $$ v_{i y}=\sqrt{-2\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(20 \mathrm{~m})}=20 \mathrm{~m} / \mathrm{s} $$ Alternative Method You can check your result using the fact that the peak altitude is given by Eq. (2.9); that is, \(y_{p}=-v_{i}^{2} / 2 s\), and so \(v_{i}^{2}=-2 g_{y}\), or \(v_{i}^{2}=-2(-9,81 \mathrm{~m} / 3)(20\) and \(u_{i}=19.8 \mathrm{~m} / \mathrm{s}\), or to two significant figures, \(u_{i}=20 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
The stone was thrown with an initial speed of 20 m/s.

Step by step solution

01

Identify Known Values

The given values are the maximum height reached, which is \( y = 20 \) m, the acceleration due to gravity \( a = -9.81 \text{ m/s}^2 \) (negative because it is downward), and the final velocity at the peak, which is \( v_{fy} = 0 \text{ m/s} \).
02

Select the Appropriate Formula

Use the kinematic equation for acceleration: \( v_{fy}^2 = v_{iy}^2 + 2ay \), where \( v_{iy} \) is the initial velocity we need to find.
03

Rearrange the Formula for Initial Velocity

Since we want to find \( v_{iy} \), rearrange the formula as \( v_{iy}^2 = v_{fy}^2 - 2ay \).
04

Substitute Known Values

Substitute the known values into the equation: \( v_{iy}^2 = 0^2 - 2(-9.81 \text{ m/s}^2)(20 \text{ m}) \).
05

Solve for Initial Velocity

Calculate \( v_{iy}^2 = 2(9.81)(20) = 392 \). Then, find \( v_{iy} = \sqrt{392} = 19.8 \text{ m/s} \). Round this to two significant figures to get \( v_{iy} = 20 \text{ m/s} \).
06

Check Using Alternative Method

Verify by using the alternate equation from the problem: \( y_p = -\frac{v_{iy}^2}{2a} \). Rearrange to get \( v_{iy}^2 = -2a y_p \). Substitute the values: \( v_{iy}^2 = -2(-9.81)(20) \) and calculate \( v_{iy} = 19.8 \text{ m/s} \), confirming the calculated speed.

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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 fundamental in the study of motion, enabling us to describe the movement of objects with precision. They relate the parameters of motion such as initial velocity, final velocity, acceleration, time, and displacement. In this scenario, we utilize these equations to determine the initial velocity of a stone thrown upward. This helps us connect different motion parameters. One of the key kinematic equations used here is:
  • \( v_{fy}^2 = v_{iy}^2 + 2ay \)
Where:\( v_{fy} \) is the final velocity, \( v_{iy} \) is the initial velocity, \( a \) is the acceleration, and \( y \) is the displacement.
These equations are highly useful because they allow us to solve for unknown variables when others are provided. They are particularly handy in tasks like determining how long it takes for an object to hit the ground or reach a certain height. Use these equations to bridge the gaps in motion understanding, making them indispensable tools in physics.
Acceleration Due to Gravity
Acceleration due to gravity, usually denoted by \( g \), is the rate at which an object accelerates when it is in free fall near the Earth's surface. The standard value of \( g \) is approximately \(-9.81 \text{ m/s}^2\). The negative sign indicates that gravity acts downward, which is crucial to understand, especially when positive upward directions are assigned, as in this problem.
The acceleration due to gravity remains constant irrespective of the object's mass. This uniformity allows us to calculate projectile motion effectively. It plays a critical role in shaping the stone's trajectory after being thrown since it continuously pulls the stone back towards the ground.
Initial Velocity Calculation
Calculating initial velocity is a crucial task in projectile motion problems. Initial velocity denotes the speed and direction an object begins its movement. To find this, we rearrange the kinematic equation appropriately, which was originally:
  • \( v_{fy}^2 = v_{iy}^2 + 2ay \)
Rearranging gives us:
  • \( v_{iy}^2 = v_{fy}^2 - 2ay \)
Substituting known values for the stone's motion, such as \( y = 20 \text{ m} \) and \( a = -9.81 \text{ m/s}^2 \), helps in solving for \( v_{iy} \). Calculating the square root allows us to determine that the initial velocity is about \( 20 \text{ m/s} \). This calculation confirms the necessary speed with which the stone was thrown upward to reach its peak height.
Free Fall Motion
Free fall motion describes the phenomenon when the only force acting on an object is gravity. This motion occurs under the acceleration due to gravity without interference from air resistance or other forces. During the stone's ascent, it initially moves against gravity, which gradually slows it down until its upward velocity becomes zero.
At the peak height of 20 meters, the velocity is zero momentarily before the stone enters free fall on its descent back to Earth. Understanding free fall is essential for accurately analyzing all stages of projectile motion, especially in determining how gravitational pull influences the stone's travel from throw to landing.

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Most popular questions from this chapter

A skier starts from rest and slides down a mountainside along a straight descending path \(9.0 \mathrm{~m}\) long in \(3.0 \mathrm{~s}\). In what time after starting will the skier acquire a speed of \(24 \mathrm{~m} / \mathrm{s}\) ? Assume that the acceleration is constant and the entire run is straight, at a fixed incline, and around \(1.0\) -km long. We must find the skier's acceleration from the data concerning the 3.0 s trip. Taking the direction of motion down the inclined path as the \(+x\) -direction, we have \(t=3.0 \mathrm{~s}, v_{i x}=0\), and \(x=9.0 \mathrm{~m} .\) Then \(x=v_{i x} t+\frac{1}{2} a t^{2}\) gives $$ a=\frac{2 x}{t^{2}}=\frac{18 \mathrm{~m}}{(3.0 \mathrm{~s})^{2}}=2.0 \mathrm{~m} / \mathrm{s}^{2} $$ We can now use this value of \(a\) for the longer trip, from the starting point to the place where \(v_{f x}=24 \mathrm{~m} / \mathrm{s}\). For this trip, \(v_{i x}=0\), \(u_{f x}=24 \mathrm{~m} / \mathrm{s}, a=2.0 \mathrm{~m} / \mathrm{s}^{2} .\) Then, from \(v_{f}=v_{i}+a t\) $$ t=\frac{v_{f x}-v_{i x}}{a}=\frac{24 \mathrm{~m} / \mathrm{s}}{2.0 \mathrm{~m} / \mathrm{s}^{2}}=12 \mathrm{~s} $$

A drone on a runway accelerates from rest at a constant rate of \(4.00\) \(\mathrm{m} / \mathrm{s}^{2}\). It travels \(20.0 \mathrm{~m}\) before lifting off the ground. What speed did it attain as it became airborne? [Hint: You are given \(u_{i}, a\), and s, and you need to find uf.]

People working for National Geographic dropped a peregrine falcon from a plane at an altitude of \(4572 \mathrm{~m}(15000 \mathrm{ft})\). The bird dove down reaching a speed of about \(81.8 \mathrm{~m} / \mathrm{s}(183 \mathrm{mph})\). Determine its acceleration assuming it to be constant. [Hint: The bird was dropped, not thrown down.]

A ball is thrown upward at an angle of \(30^{\circ}\) to the horizontal and lands on the top edge of a building that is \(20 \mathrm{~m}\) away. The top edge is \(5.0 \mathrm{~m}\) above the throwing point. How fast was the ball thrown?

An object starts from rest with a constant acceleration of \(8.00 \mathrm{~m} / \mathrm{s}^{2}\) along a straight line. Find \((a)\) the speed at the end of \(5.00 \mathrm{~s},(b)\) the average speed for the 5-s interval, and ( \(c\) ) the distance traveled in the \(5.00 \mathrm{~s}\). We are interested in the motion for the first \(5.00 \mathrm{~s}\). Take the direction of motion to be the \(+x\) -direction (that is, \(s=x\) ). We know that \(v_{i}=0, t=5.00 \mathrm{~s}\), and \(a=8.00 \mathrm{~m} / \mathrm{s}^{2}\). Because the motion is uniformly accelerated, the five motion equations apply. (a) \(v_{f x}=v_{i x}+a t=0+\left(8.00 \mathrm{~m} / \mathrm{s}^{2}\right)(5.00 \mathrm{~s})=40.0 \mathrm{~m} / \mathrm{s}\) (b) \(\quad v_{a v}=\frac{v_{i x}+v_{f x}}{2}=\frac{0+40.0}{2} \mathrm{~m} / \mathrm{s}=20.0 \mathrm{~m} / \mathrm{s}\) (c) \(x=v_{i x} t+\frac{1}{2} a t^{2}=0+\frac{1}{2}\left(8.00 \mathrm{~m} / \mathrm{s}^{2}\right)(5.00 \mathrm{~s})^{2}=100 \mathrm{~m} \quad\) or \(\quad x=v_{a v} t=(20.0 \mathrm{~m} / \mathrm{s})(5.00 \mathrm{~s})=100 \mathrm{~m}\)
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