/*! 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 99 A ball is thrown vertically down... [FREE SOLUTION] | 91Ó°ÊÓ

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A ball is thrown vertically downward from the top of a 36.6m-tall building. The ball passes the top of a window that is \(12.2 \mathrm{~m}\) above the ground \(2.00 \mathrm{~s}\) after being thrown. What is the speed of the ball as it passes the top of the window?

Short Answer

Expert verified
The speed of the ball as it passes the window is 22.0 m/s.

Step by step solution

01

Analyze the Problem

We need to find the speed of the ball as it passes a window located 12.2 meters above the ground level, given that the ball was thrown from a height of 36.6 meters. We also know that it reaches this window 2 seconds after being released.
02

Understand the Physics Involved

Since the ball is thrown downward, we use a kinematic equation. The equations of motion relate initial velocity, time, acceleration, and displacement. Here, acceleration is due to gravity (approximately 9.8 m/s²).
03

Choose the Correct Kinematic Equation

The kinematic equation that relates initial velocity, time, acceleration, and displacement is \[ s = ut + \frac{1}{2}at^2 \] where \( s \) is the displacement, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.
04

Calculate Displacement

The displacement \( s \) is the difference between the initial height (36.6 m) and the window's height (12.2 m), \[ s = 36.6 \, \text{m} - 12.2 \, \text{m} = 24.4 \, \text{m} \]
05

Set Up Equation with Known Values

Insert the known values into the kinematic equation:\[ 24.4 = u \times 2 + \frac{1}{2} \times 9.8 \times (2)^2 \]
06

Solve for the Initial Velocity

Simplify the equation:\[ 24.4 = 2u + \frac{1}{2} \times 9.8 \times 4 \]\[ 24.4 = 2u + 19.6 \]Solve for \( u \):\[ 2u = 24.4 - 19.6 \]\[ 2u = 4.8 \]\[ u = \frac{4.8}{2} = 2.4 \, \text{m/s} \]
07

Determine the Final Velocity

Now use the formula \( v = u + at \) to find the final speed:\[ v = 2.4 + 9.8 \times 2 \]\[ v = 2.4 + 19.6 = 22.0 \, \text{m/s} \]

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Key Concepts

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

Kinematic Equations
Understanding projectile motion involves grasping the kinematic equations. These are mathematical formulas that describe the motion of objects under uniform acceleration. They help us understand how objects like the ball in our problem move over time. Essential equations include:
  • Displacement, initial velocity, acceleration, and time relation: \[ s = ut + \frac{1}{2}at^2 \]
  • Final velocity, initial velocity, acceleration, and time relation: \[ v = u + at \]
These equations assume that the only force acting on an object is a constant force, such as gravity. In our exercise, the ball is thrown downward, making these equations perfect for calculating its movement. Each term in the equations holds importance: - **\(s\)** represents displacement, which is how far the object has traveled.- **\(u\)** stands for initial velocity, or the speed the object begins with.- **\(a\)** is the constant acceleration, here due to gravity.- **\(t\)** is the time over which the motion occurs. Whether it involves finding out how fast something moves or how far it travels, these equations are the go-to tools for analyzing linear motion.
Acceleration Due to Gravity
In the context of projectile motion, gravity plays a crucial role, providing a constant acceleration that impacts the object's speed and trajectory. On Earth, the acceleration due to gravity is approximately \(9.8 \, \text{m/s}^2\). This constant value simplifies calculations since it remains uniform throughout the motion of the object.
  • It always acts downwards, towards the Earth's center, affecting the vertical component of the motion.
  • It influences how quickly an object's velocity changes over time.
In our exercise, as the ball is thrown downward, the acceleration due to gravity increases its velocity over time. This means that after 2 seconds, the ball moves faster than when it was released. Remember, gravity efficiently transforms potential energy into kinetic energy, affecting both falling and rising objects. Understanding its constant and unidirectional nature helps in precisely determining the velocity and position of moving objects like our ball.
Initial Velocity Calculation
Calculating an object's initial velocity is an important step in understanding its motion. This velocity is the speed at which the object begins its journey. To find it, we can use the kinematic equation:\[ s = ut + \frac{1}{2}at^2 \] where all variables come into play.
  • **\(s\)** is the total vertical displacement; in our problem, it's from the top of the building to just above the window.
  • **\(t\)** is the time taken to reach the window.
  • **\(a\)** is the acceleration due to gravity.
Plugging these values into the formula allows us to isolate \(u\), the initial velocity. In our exercise, we rearrange the equation to find:\[ u = \frac{s - \frac{1}{2}at^2}{t} \] substituting known values gives \(2.4 \, m/s\). This calculated value helps in predicting future positions and speeds of the ball during its descent. Mastering how to derive initial velocity aids in a clearer comprehension of the ball's overall motion.

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Most popular questions from this chapter

A certain juggler usually tosses balls vertically to a height \(H\). To what height must they be tossed if they are to spend twice as much time in the air?

When startled, an armadillo will leap upward. Suppose it rises \(0.544 \mathrm{~m}\) in the first \(0.200 \mathrm{~s}\). (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of \(0.544 \mathrm{~m}\) ? (c) How much higher does it go?

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(a) With what speed must a ball be thrown vertically from ground level to rise to a maximum height of \(50 \mathrm{~m}\) ? (b) How long will it be in the air? (c) Sketch graphs of \(y, v\), and \(a\) versus \(t\) for the ball. On the first two graphs, indicate the time at which \(50 \mathrm{~m}\) is reached.

The sport with the fastest moving ball is jai alai, where measured speeds have reached \(303 \mathrm{~km} / \mathrm{h}\). If a professional jai alai player faces a ball at that speed and involuntarily blinks, he blacks out the scene for \(100 \mathrm{~ms}\). How far does the ball move during the blackout?

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