/*! 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 101 A ball is thrown straight up. It... [FREE SOLUTION] | 91Ó°ÊÓ

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A ball is thrown straight up. It passes a 2.00-m-high window \(7.50 \mathrm{m}\) off the ground on its path up and takes \(1.30 \mathrm{s}\) to go past the window. What was the ball's initial velocity?

Short Answer

Expert verified
The ball's initial velocity is approximately \(6.62 \, \text{m/s}\).

Step by step solution

01

Identify the given values

We are given: - Height of the window: \(h = 2.00 \,\text{m}\) - Time taken to go past the window: \(t = 1.30 \, \text{s}\) - Acceleration due to gravity: \(a = -9.81 \, m/s^2\) (downward)
02

Calculate the distance traveled during the given time

Using the equation of motion: \[d = v_i*t + \frac{1}{2} * a*t^2\] We know the distance traveled is equal to the height of the window. Therefore, we can substitute the given values in the equation: \[2.00 \,\text{m} = v_i * 1.30 \,\text{s} + \frac{1}{2} * (-9.81 \,\text{m/s}^2) * (1.30 \,\text{s})^2\]
03

Solve the equation for the initial velocity

First, we will solve the equation in the parentheses: \[\frac{1}{2} * (-9.81 \,\text{m/s}^2) * (1.30 \,\text{s})^2 = -6.61145 \,\text{m}\] Now, we will substitute this value back into the equation: \[2.00 \,\text{m} = v_i * 1.30 \,\text{s} - 6.61145 \,\text{m}\] Next, add 6.61145 m to both sides of the equation: \[8.61145 \,\text{m} = v_i * 1.30 \,\text{s}\] Finally, divide both sides by 1.30 s to find the initial velocity: \[v_i = \frac{8.61145 \,\text{m}}{1.30 \,\text{s}} = 6.62 \, \text{m/s}\] The ball's initial velocity is approximately \(6.62 \, \text{m/s}\).

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Key Concepts

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

Initial Velocity
Understanding the initial velocity is crucial for solving many kinematics problems. It is the speed at which an object starts its motion. Importantly, velocity is a vector, which means it has both magnitude (speed) and direction. In the context of the ball being thrown upwards, the initial velocity (\(v_i\)) is the speed at which it leaves the thrower's hand.

When dealing with problems involving objects thrown vertically, the initial velocity can determine how high the object will go and how long it will stay in the air before it starts to fall back down. The calculation of initial velocity often includes knowing the distance traveled and the acceleration acting on the object, usually gravity in vertically thrown objects. It serves as a starting point for analyzing the motion.
Equation of Motion
Equations of motion are a set of formulas that relate displacement, initial velocity, final velocity, acceleration, and time in problems involving constant acceleration. A common equation of motion used in kinematics problem solving is:
\[d = v_i*t + \frac{1}{2} * a*t^2\]
where \(d\) stands for displacement, \(v_i\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time.

In the case of the ball passing the window, this specific equation allows us to calculate how far it traveled within a certain time frame while under constant acceleration. It's worth noting that the motion being analyzed here is in one dimension (vertically up and down), which simplifies the use of the equation.
Acceleration Due to Gravity
The acceleration due to gravity is a pivotal concept in kinematics. It represents the constant rate at which objects accelerate towards Earth when in free fall, and it is denoted by \(g\), typically with a value of approximately \(-9.81 \, m/s^2\) on the surface of the Earth.

The negative sign indicates that gravity acts downwards, toward the center of the Earth. This acceleration impacts the motion of any object moving under the influence of Earth's gravity, such as the thrown ball in our problem. It's worth emphasizing that this value remains constant near the Earth's surface regardless of the object's mass.
Physics Equations
Physics equations are the tools that quantify and explain the natural phenomena happening around us. They can be as broad as Einstein's equation of energy equivalence, \(E=mc^2\), to as specific as the kinematic equations used for objects in motion under constant acceleration.

These equations are derived from empirical observations and mathematical reasoning, forming the framework through which we understand the physical world. In kinematics problem solving, they enable us to predict future motion or reconstruct the history of an object's movement from limited information. Mastery of physics equations and understanding their application is fundamental in accurately solving problems and making sense of the forces that govern the motion of objects in our universe.

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Most popular questions from this chapter

A woodpecker's brain is specially protected from large accelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker's head comes to a stop from an initial velocity of \(0.600 \mathrm{m} / \mathrm{s}\) in a distance of only \(2.00 \mathrm{mm}\). (a) Find the acceleration in meters per second squared and in multiples of \(g\), where \(g=9.80 \mathrm{m} /\) s \(^{2}\). (b) Calculate the stopping time. (c) The tendons cradling the brain stretch, making its stopping distance \(4.50 \mathrm{mm}\) (greater than the head and, hence, less acceleration of the brain). What is the brain's acceleration, expressed in multiples of \(g\) ?

A particle has a constant acceleration of \(6.0 \mathrm{m} / \mathrm{s}^{2}\) (a) If its initial velocity is \(2.0 \mathrm{m} / \mathrm{s}\), at what time is its displacement 5.0 m? (b) What is its velocity at that time?

Compare the distance traveled of an object that undergoes a change in velocity that is twice its initial velocity with an object that changes its velocity by four times its initial velocity over the same time period. The accelerations of both objects are constant.

A 10.0-m-long truck moving with a constant velocity of \(97.0 \mathrm{km} / \mathrm{h}\) passes a 3.0 -m-long car moving with a constant velocity of \(80.0 \mathrm{km} / \mathrm{h}\). How much time elapses between the moment the front of the truck is even with the back of the car and the moment the back of the truck is even with the front of the car?

Calculate the displacement and velocity at times of (a) \(0.500 \mathrm{s},\) (b) \(1.00 \mathrm{s},\) (c) \(1.50 \mathrm{s},\) and \((\mathrm{d}) 2.00 \mathrm{s}\) for a ball thrown straight up with an initial velocity of \(15.0 \mathrm{m} / \mathrm{s}\). Take the point of release to be \(y_{0}=0\).

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