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What is the velocity at the midway point of a ball able to reach a height \(y\) when thrown with an initial velocity \(v_{0} ?\)

Short Answer

Expert verified
Answer: The velocity at the midway point of the ball is \(\frac{v_{0}}{2}\).

Step by step solution

01

Identify the equations of motion

We will use the following equations of motion to solve the problem: 1. \(v=v_{0}+at\) 2. \(y=v_{0}t+\frac{1}{2}at^2\) Here, \(v\) is the final velocity of the ball, \(v_{0}\) is the initial velocity, \(a\) is the acceleration due to gravity, \(t\) is the time taken to reach that height y, and \(y\) is the height reached by the ball.
02

Determine the height at the midway point

Since the midway point corresponds to half the maximum height of the ball, we will consider the height \(y\) to be \(\frac{1}{2}y_{max}\). So let's denote it as: \(h = \frac{1}{2}y_{max}\)
03

Calculate the maximum height reached

To find the maximum height reached by the ball, we can use the equation of motion when the ball's velocity at the maximum height becomes 0: \(0^2 = v_{0}^2 - 2ay_{max}\) Solving for \(y_{max}\), we get: \(y_{max} = \frac{v_{0}^2}{2a}\), where \(a = -g\), the acceleration due to gravity. (We consider it negative since it's acting opposite to the direction of the throw)
04

Determine the time taken to reach midway point

To find the time taken to reach the midway point, we will use the height at the midway point (h) and the second equation of motion: \(h = v_{0}t+\frac{1}{2}at^2\) Substitute \(h\) with \(\frac{1}{2}y_{max}\) and \(y_{max}\) with \(\frac{v_{0}^2}{2a}\): \(\frac{1}{2}\times\frac{v_{0}^2}{2a} = v_{0}t+\frac{1}{2}at^2\) Solving for t, we get: \(t = \frac{v_{0}}{2a}\)
05

Calculate the velocity at the midway point

Now that we have the time taken to reach the midway point, we can find the velocity at this point using the first equation of motion. \(v = v_{0} + at\) Plug in the values of \(t\) and \(a\): \(v = v_{0} - g\frac{v_{0}}{2g}\) \(v = \frac{v_{0}}{2}\) So, the velocity at the midway point of a ball able to reach a height \(y\) when thrown with an initial velocity \(v_{0}\) is \(\frac{v_{0}}{2}\).

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Key Concepts

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

Velocity
Velocity is a key concept in kinematics and refers to the rate at which an object changes its position. It's a vector quantity, meaning it has both magnitude and direction. Understanding the velocity of a moving object helps us predict its future position and calculate its energy.
  • Initial Velocity \((v_0)\): This is the speed and direction of an object at the beginning of its motion. In our context, it is the speed at which the ball is thrown.
  • Final Velocity \((v)\): This is the speed of the object at a particular point in time or position. In the problem, we calculated the velocity at the midway point of the motion.
In projectile motion, the velocity changes due to the effect of gravity. Understanding velocity allows us to apply the equations of motion to find various parameters of a moving object, such as time of flight and maximum height.
Projectile motion
Projectile motion describes the motion of an object thrown into the air, where only the force acting on it is gravity. This motion has two components, horizontal and vertical, which are analyzed separately. When a ball is thrown, it follows a parabolic trajectory due to these components.
In this context:
  • The horizontal motion is uniform, meaning there's no acceleration because air resistance is typically negligible.
  • The vertical motion, however, is influenced by gravity, causing acceleration downward.
The key to solving many problems in projectile motion is separating these components and applying the equations of motion independently. At the midway point, the ball's vertical velocity has decreased because of gravity, affecting both its height and speed.
Equations of motion
Equations of motion are mathematical formulas that describe the behavior of moving objects. They connect concepts such as velocity, acceleration, time, and displacement. In solving kinematic problems, these equations help predict how objects will move under different forces. The two primary equations used in our exercise are:
  • \(v = v_0 + at\) : This equation provides a direct relation between velocity, initial velocity, acceleration, and time.
  • \(y = v_0 t + \frac{1}{2} at^2\) : This equation relates displacement, initial velocity, time, and acceleration.
In our problem, we used these equations to find both the time it takes for the ball to reach the midway point and the velocity at that point. By understanding these equations, you can predict an object's position or velocity at any given time during its motion. These equations form the foundation for analyzing any type of motion in physics.

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Most popular questions from this chapter

An object is thrown upward with a speed of \(28.0 \mathrm{~m} / \mathrm{s}\). How high above the projection point is it after 1.00 s?

A car traveling at \(25.0 \mathrm{~m} / \mathrm{s}\) applies the brakes and decelerates uniformly at a rate of \(1.2 \mathrm{~m} / \mathrm{s}^{2}\) a) How far does it travel in \(3.0 \mathrm{~s}\) ? b) What is its velocity at the end of this time interval? c) How long does it take for the car to come to a stop? d) What distance does the car travel before coming to a stop?

The position of an object as a function of time is given as \(x=A t^{3}+B t^{2}+C t+D .\) The constants are \(A=2.1 \mathrm{~m} / \mathrm{s}^{3}\) \(B=1.0 \mathrm{~m} / \mathrm{s}^{2}, C=-4.1 \mathrm{~m} / \mathrm{s},\) and \(D=3 \mathrm{~m}\) a) What is the velocity of the object at \(t=10.0 \mathrm{~s}\) ? b) At what time(s) is the object at rest? c) What is the acceleration of the object at \(t=0.50 \mathrm{~s} ?\) d) Plot the acceleration as a function of time for the time interval from \(t=-10.0 \mathrm{~s}\) to \(t=10.0 \mathrm{~s}\).

You are trying to improve your shooting skills by shooting at a can on top of a fence post. You miss the can, and the bullet, moving at \(200 . \mathrm{m} / \mathrm{s},\) is embedded \(1.5 \mathrm{~cm}\) into the post when it comes to a stop. If constant acceleration is assumed, how long does it take for the bullet to stop?

An object is thrown upward with a speed of \(28.0 \mathrm{~m} / \mathrm{s}\). How long does it take it to reach its maximum height?

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