/*! 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 52 A ball is thrown straight upward... [FREE SOLUTION] | 91Ó°ÊÓ

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A ball is thrown straight upward. At \(4.00 \mathrm{m}\) above its launch point, the ball's speed is one-half its launch speed. What maximum height above its launch point does the ball attain?

Short Answer

Expert verified
The maximum height above the launch point is 12.00 m.

Step by step solution

01

Understanding the Problem

We are given that a ball is thrown upwards and when it is 4.00 m above its launch point, its speed is half of its initial launch speed. We need to find the maximum height the ball reaches above the launch point.
02

Using Energy Conservation Principle

According to the conservation of mechanical energy, the sum of kinetic and potential energy at the launch point must be equal to the sum at any other point in its motion, including the point of maximum height.
03

Set Up the Conservation Equation

Let the launch speed be \(v_i\). At launch, the kinetic energy is \(\frac{1}{2} m v_i^2\) and potential energy is 0. At the point where the ball is 4.00 m above the launch point, the speed is \(\frac{1}{2}v_i\), so the kinetic energy is \(\frac{1}{2} m \left(\frac{v_i}{2}\right)^2\) and the potential energy is \(mg \times 4.00\).
04

Calculate Initial Speed Using Known Conditions

Set the total mechanical energy at the start equal to the total mechanical energy 4.00 m up: \( \frac{1}{2} m v_i^2 = \frac{1}{2} m \left(\frac{v_i}{2}\right)^2 + mg \times 4.00 \). Solve this equation for the initial speed \(v_i\).
05

Solve for Maximum Height

At maximum height, all kinetic energy will have been converted into potential energy, since speed is 0 at the top. Use \( \frac{1}{2} m v_i^2 = mgh_{max} \) to solve for \(h_{max}\), where \(g\) is the acceleration due to gravity (9.8 m/s²).
06

Calculate the Maximum Height

With \(v_i^2 = 2g \times 4.00\), use \( h_{max} = \frac{v_i^2}{2g} \) to find the total height reached. From this, subtract 4.00 m to find the maximum height above the launch point.

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Key Concepts

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

Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the causes of the motion, such as forces or energy. When solving problems involving motion, kinematics provides a foundation by using equations that describe the velocity, acceleration, and displacement of an object. One of the key kinematic equations that is particularly useful in projectile motion is the relation between velocity, acceleration, and displacement:
  • Given initial velocity (\( v_i \))
  • Acceleration (typically due to gravity, \( g = 9.8 \,\text{m/s}^2 \))
  • Displacement (\( s \)), we have \( v^2 = v_i^2 + 2as \)
In the given problem about a ball being thrown upwards, kinematics helps determine how the change in speed relates to the position of the ball. Initially, the ball is at maximum speed, slowing down due to gravity until its speed becomes zero at its peak height. Understanding how to express these changes using kinematic equations is crucial in calculating various aspects of the ball's trajectory.
Conservation of Energy
The principle of conservation of energy states that the total energy of an isolated system remains constant. In the context of the ball thrown upward, this conservation is observed between kinetic and potential energies. The ball possesses kinetic energy due to its motion and potential energy due to its elevation above the ground. These energies convert into one another but their sum remains constant.
  • Kinetic Energy: \( KE = \frac{1}{2} m v^2 \)
  • Potential Energy: \( PE = mgh \)
Initially, the ball has no potential energy (at ground level) but maximum kinetic energy. As it rises, the kinetic energy decreases (since speed decreases) converting into potential energy (since elevation increases). The exercise uses this principle to equate the energies at different stages of the ball's motion to find out the initial speed and to calculate the maximum height reached. By equating the energies at the start and the peak (where kinetic energy becomes zero at maximum height), one can solve for how high the ball travels.
Projectile Motion
Projectile motion is a type of motion experienced by an object moving through space influenced only by gravity. When dealing with projectile motion, you primarily deal with two components of motion: the horizontal and the vertical. In this particular vertical-only motion, as the ball goes upwards, the most interesting aspect is how gravity affects the motion directly:
  • Vertical Motion: Affected by gravity, the ball's vertical speed decreases until it reaches the peak, then increases as it comes back down.
  • Acceleration Due to Gravity: This consistent force causes the change in motion, always acting downwards with magnitude \( g = 9.8 \,\text{m/s}^2 \).
  • Maximum Height: The velocity becomes zero momentarily at the peak of its trajectory.
The problem considers the ball as a projectile moving straight up, simplifying our calculations by focusing solely on vertical motion. By using the details of gravitational influence and initial conditions, you can compute how high the projectile rises when thrown upwards. This understanding aligns perfectly with the calculations made using kinematics and energy conservation principles in the exercise.

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Most popular questions from this chapter

Over a time interval of 2.16 years, the velocity of a planet orbiting a distant star reverses direction, changing from \(+20.9 \mathrm{km} / \mathrm{s}\) to \(-18.5 \mathrm{km} / \mathrm{s} .\) Find (a) the total change in the planet's velocity (in \(\mathrm{m} / \mathrm{s}\) ) and (b) its average acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ) during this interval. Include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration.

A motorcycle has a constant acceleration of \(2.5 \mathrm{m} / \mathrm{s}^{2} .\) Both the velocity and acceleration of the motorcycle point in the same direction. How much time is required for the motorcycle to change its speed from. (a) 21 to \(31 \mathrm{m} / \mathrm{s},\) and (b) 51 to \(61 \mathrm{m} / \mathrm{s} ?\)

A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of \(86.0 \mathrm{m} / \mathrm{s}^{2}\) for 1.70 seconds, at which point its fuel abruptly runs out. Air resistance has no effect on its flight. What maximum altitude (above the ground) will the rocket reach?

A hot-air balloon is rising straight up at a constant speed of \(7.0 \mathrm{m} / \mathrm{s} .\) When the balloon is \(12.0 \mathrm{m}\) above the ground, a gun fires a pellet straight up from ground level with an initial speed of \(30.0 \mathrm{m} / \mathrm{s}\). Along the paths of the balloon and the pellet, there are two places where each of them has the same altitude at the same time. How far above ground are these places?

The initial velocity and acceleration of four moving objects at a given instant in time are given in the following table. Determine the final speed of each of the objects, assuming that the time elapsed since \(t=0 \mathrm{s}\) is \(2.0 \mathrm{s}\). $$ \begin{array}{lcc} \hline & \text { Initial velocity } v_{0} & \text { Acceleration } a \\ \hline \text { (a) } & +12 \mathrm{m} / \mathrm{s} & +3.0 \mathrm{m} / \mathrm{s}^{2} \\ \text { (b) } & +12 \mathrm{m} / \mathrm{s} & -3.0 \mathrm{m} / \mathrm{s}^{2} \\\ \text { (c) } & -12 \mathrm{m} / \mathrm{s} & +3.0 \mathrm{m} / \mathrm{s}^{2} \\\ \text { (d) } & -12 \mathrm{m} / \mathrm{s} & -3.0 \mathrm{m} / \mathrm{s}^{2} \\\ \hline \end{array} $$

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