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Chapter 2: Problem 63

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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Most popular questions from this chapter

An object starts from rest and has an acceleration given by \(a=B t^{2}-\frac{1}{2} C t,\) where \(B=2.0 \mathrm{~m} / \mathrm{s}^{4}\) and \(C=-4.0 \mathrm{~m} / \mathrm{s}^{3}\). a) What is the object's velocity after 5.0 s? b) How far has the object moved after \(t=5.0\) s?

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