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Chapter 2: Problem 50

A ball is thrown straight upward with a speed \(u\) from a point \(h\) meters above the ground. Show that the time taken for the ball to strike the ground \((v / g)\left[1+\sqrt{1+\left(2 h g / v^{2}\right)}\right]\) where \(g\) is positive.

Short Answer

Expert verified
The time taken is \(\frac{v}{g} \left(1 + \sqrt{1 + \frac{2gh}{v^2}}\right)\)."

Step by step solution

01

Identify Initial Parameters

We begin with the initial conditions of the problem. The ball is thrown upwards with velocity \(u\) from a height \(h\) and will eventually reach the ground. This involves motion under gravity \(g\).
02

Write the Motion Equation

Using the equation of motion: \(s = ut + \frac{1}{2}at^2\), where \(s\) is the displacement, \(a = -g\) (since gravity acts downward), and the displacement \(s = -h\) because it falls from height \(h\) to the ground. This can be rewritten as \(-h = ut - \frac{1}{2}gt^2\).
03

Rearrange to a Standard Quadratic Form

Rearrange the motion equation to fit the standard quadratic form \(at^2 + bt + c = 0\). Thus, we have \(\frac{1}{2}gt^2 - ut - h = 0\).
04

Apply the Quadratic Formula

The quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) applies to our equation. Here, \(a = \frac{1}{2}g\), \(b = -u\), and \(c = -h\). Substitute these into the formula: \[t = \frac{-(-u) \pm \sqrt{(-u)^2 - 4(\frac{1}{2}g)(-h)}}{2(\frac{1}{2}g)}\].
05

Simplify the Expression

Simplify the quadratic solution: \[t = \frac{u \pm \sqrt{u^2 + 2gh}}{g}\]. Since time cannot be negative, use the positive root: \[t = \frac{u + \sqrt{u^2 + 2gh}}{g}\].
06

Express in Required Form

Recognize that \(u\) is given as \(v\) in the required expression. The solution is expressed in the form \(\frac{v}{g} \left(1 + \sqrt{1 + \frac{2gh}{v^2}}\right)\). On examining the expression, it's observed that it matches once the substitution \(u = v\) is made.

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Key Concepts

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

Equations of Motion
In kinematics, the equations of motion are pivotal for analyzing objects moving within a straight path, especially under constant acceleration. When a ball is thrown upwards, it follows a path described by these equations, influenced by Earth's gravity pulling it down. The primary equation used here is:
  • \(s = ut + \frac{1}{2}at^2\)
This represents the displacement \(s\) of an object thrown upwards with initial velocity \(u\) and experiences constant acceleration \(a\) over time \(t\). Here, gravity is the acceleration \(a\), acting downwards. Therefore, in our exercise:
  • The displacement \(s\) is \(-h\) as the ball goes from height \(h\) to the ground.
  • The acceleration \(a = -g\).
This is transformed into \(-h = ut - \frac{1}{2}gt^2\). Recognizing and rearranging equations of motion is essential for solving kinematic problems. They help link variables like initial velocity, time, acceleration, and displacement into one comprehensive framework.
Gravity
Gravity is a powerful force acting on all objects with mass on Earth, pulling them toward its center. It's crucial in kinematic equations as it dictates constant acceleration experienced by every object in free fall. This exercise demonstrates gravity's role explicitly. Here are key points to understand:
  • Gravity's constant acceleration, denoted as \(g\), is approximately 9.8 m/s虏 on Earth.
  • It acts downwards, influencing the vertical motion of objects.
  • In our equations, gravity is negative when an object moves upwards, as it opposes the initial direction.
In the exercise, when the ball is thrown upwards, gravity decelerates it until it changes direction and starts falling back to the ground. Understanding gravity's influence enables solving complex vertical motion problems and assists in determining elements like time of flight and maximum height. These attributes make gravity foundational in kinematic problems.
Quadratic Formula
The quadratic formula is quintessential for solving polynomial equations of the second degree, which appear frequently in motion problems like this one. It provides solutions for equations that can be rearranged into a standard form, \(at^2 + bt + c = 0\). Understanding the quadratic formula's role is crucial:
  • The formula is expressed as \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
  • It determines the time \(t\) the ball takes to hit the ground by solving the second-degree polynomial for \(t\).
In the problem, substituting values from the rearranged motion equation \(\frac{1}{2}gt^2 - ut - h = 0\) into the quadratic formula yields the time taken for the ball鈥檚 journey to ground level. The calculation results in a solution with two possible values for \(t\), but only the positive value is meaningful in physical terms (as time cannot be negative). This application of the quadratic formula simplifies complex time measurements and is a vital tool in solving motion-related equations.

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Most popular questions from this chapter

A ball is thrown upward at an angle of \(30^{\circ}\) to the horizontal and lands on the top edge of a building that is \(20 \mathrm{~m}\) away. The top edge is \(5.0 \mathrm{~m}\) above the throwing point. How fast was the ball thrown?

A stone is thrown straight upward and it rises to a maximum height of \(20 \mathrm{~m}\). With what speed was it thrown? Take up as the positive \(y\) -direction. The stone's velocity is zero at the top of its path. Then \(v_{f y}=0, y=20 \mathrm{~m}, a=-9.81 \mathrm{~m} / \mathrm{s}^{2}\). (The minus sign arises because the acceleration due to gravity is always downward and we have taken up to be positive.) Use \(v_{s}^{2}=v_{i}^{2}+2 a y\) to find $$ v_{i y}=\sqrt{-2\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(20 \mathrm{~m})}=20 \mathrm{~m} / \mathrm{s} $$ Alternative Method You can check your result using the fact that the peak altitude is given by Eq. (2.9); that is, \(y_{p}=-v_{i}^{2} / 2 s\), and so \(v_{i}^{2}=-2 g_{y}\), or \(v_{i}^{2}=-2(-9,81 \mathrm{~m} / 3)(20\) and \(u_{i}=19.8 \mathrm{~m} / \mathrm{s}\), or to two significant figures, \(u_{i}=20 \mathrm{~m} / \mathrm{s}\).

A robot named Fred is initially moving at \(2.20 \mathrm{~m} / \mathrm{s}\) along a hallway in a space terminal. It subsequently speeds up to \(4.80 \mathrm{~m} / \mathrm{s}\) in a time of \(0.20 \mathrm{~s}\). Determine the size or magnitude of its average acceleration along the path traveled. The defining scalar equation is \(a_{a v}=\left(v_{f}-v_{i}\right) / t\). Everything is in proper SI units, so we need only carry out the calculation: $$ a_{a v}=\frac{4.80 \mathrm{~m} / \mathrm{s}-2.20 \mathrm{~m} / \mathrm{s}}{0.20 \mathrm{~s}}=13 \mathrm{~m} / \mathrm{s}^{2} $$ Notice that the answer has two significant figures because the time has only two significant figures.

A body projected upward from the level ground at an angle of \(50^{\circ}\) with the horizontal has an initial speed of \(40 \mathrm{~m} / \mathrm{s}\). \((a)\) How long will it take to hit the ground? \((b)\) How far from the starting point will it strike? (c) At what angle with the horizontal will it strike?

If a vehicle accelerates at \(10.0 \mathrm{~m} / \mathrm{s}^{2}\) from rest for \(20.0 \mathrm{~s}\), how far will it travel in the process? [Hint: You are given \(a, u_{i}\), and \(t\), and you need to find s.]
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