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Chapter 1: Problem 1

A boy 2 meters tall shoots a toy rocket straight up from head level at 10 meters per second. Assume the acceleration of gravity is 9.8 meters/sec \(^{2}\). (a) What is the highest point above the ground reached by the rocket? (b) When does the rocket hit the ground?

Short Answer

Expert verified
The highest point above the ground reached by the rocket is approximately 7.14 meters, and the rocket hits the ground after approximately 4.87 seconds.

Step by step solution

01

Identify the given values

The initial height, \(h_0\), is 2 meters (head level of the boy). The initial velocity, \(v_0\), is 10 m/s (straight up). The acceleration due to gravity, \(g\), is 9.8 m/s虏 (downward).
02

Finding the time when the rocket reaches its highest point

At the highest point, the final velocity of the rocket, \(v\), will be 0. We can use the following kinematic equation to find the time \(t\) at which this happens: \(v = v_0 - gt\) Rearrange the equation to find the time, \(t\): \(t = \frac{v_0 - v}{g}\) Since \(v = 0\), we have: \(t = \frac{v_0}{g}\) Now, plug in the given values: \(t = \frac{10}{9.8} 鈮� 1.02 \, seconds\)
03

Finding the highest point above the ground

Now, we will use the time, \(t = 1.02 \, seconds\), we found in Step 2 to find the height above the ground (\(h\)) at this point. Use the kinematic equation and plug in the given values: \(h = h_0 + v_0t - \frac{1}{2}gt^2\) \(h = 2 + (10)(1.02) - \frac{1}{2}(9.8)(1.02)^2\) \(h 鈮� 7.14 \, meters\) So, the highest point above the ground reached by the rocket is approximately 7.14 meters.
04

Finding the time when the rocket hits the ground

Now we will use the kinematic equation with the final height, \(h\), set to 0 to find the time, \(t\), when the rocket hits the ground: \(0 = h_0 + v_0t - \frac{1}{2}gt^2\) Rearrange the equation to find the time, \(t\): \(t = \frac{-v_0 卤 \sqrt{v_0^2 - 4(-\frac{1}{2}g)(-h_0)}}{g}\) Plug in the given values: \(t = \frac{-10 卤 \sqrt{10^2 - 4(-\frac{1}{2}(9.8))(-2)}}{9.8}\) \(t 鈮� 0.42s, 4.87s\) Since the first value corresponds to the time before the rocket has reached its highest point, we will take the second value as our answer. The rocket hits the ground after approximately 4.87 seconds.

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Key Concepts

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

Motion Under Gravity
Understanding how objects move under the influence of gravity is an essential concept in physics, frequently described using kinematic equations. Imagine you're observing an apple falling from a tree. That apple accelerates towards the ground due to Earth's gravitational pull, which is constant at approximately 9.8 meters per second squared (\( g \) ).

Ideally, without air resistance, an object like our toy rocket from the exercise example would accelerate continuously as long as it's moving downwards, and its upward motion would decelerate at this same rate until it reaches the highest point, much like the apple halting its motion momentarily at the peak of its fall.
Projectile Motion
Projectile motion occurs when an object is thrown or projected into the air, moving under the influence of gravity only, while the only force applied is at the beginning, during the push or the throw. In our scenario, the toy rocket follows a vertical trajectory, propelled upwards at an initial velocity of 10 m/s. The rocket's motion can be broken down into two key components: the upward launch and the subsequent fall back to earth.

Significantly, the highest point the projectile reaches, known as the apex of its trajectory, is where the velocity becomes zero for a split second before gravity pulls it back down. Using kinematic equations helps us to calculate these crucial moments of the projectile's path, like the time it takes to hit the ground after being launched.
Quadratic Equations
Quadratic equations appear in a variety of scientific and mathematical contexts, including the analysis of projectile motion. A standard form of a quadratic equation is given by \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( x \) represents an unknown variable. In our textbook solution, we encounter a quadratic equation as part of finding out when the rocket will hit the ground, after being launched into the air.

Solving this equation usually involves finding the values of \( x \) for which the equation is true, also known as the roots of the equation. These roots can be found using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). In our case, the positive root gives us the time taken for the rocket to return to the ground, revealing two possible times, of which we take the larger one as the realistic scenario for a projectile launched upwards and then descending to the ground.

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Most popular questions from this chapter

A simple pendulum consists of a particle of mass \(m\) supported by a piece of string of length \(L .\) Assuming that the pendulum is displaced through an angle \(\theta_{0}\) radians from the vertical and then released from rest, the resulting motion is described by the initial-value problem $$\frac{d^{2} \theta}{d t^{2}}+\frac{g}{L} \sin \theta=0, \quad \theta(0)=\theta_{0}, \quad \frac{d \theta}{d t}(0)=0$$ $$(1.11 .28)$$ (a) For small oscillations, \(\theta<<1,\) we can use the approximation \(\sin \theta \approx \theta\) in Equation \((1.11 .28)\) to obtain the linear equation $$\frac{d^{2} \theta}{d t^{2}}+\frac{g}{L} \theta=0, \quad \theta(0)=\theta_{0}, \quad \frac{d \theta}{d t}(0)=0$$. Solve this initial- value problem for \(\theta\) as a function of \(t .\) Is the predicted motion reasonable? (b) Obtain the following first integral of \((1.11 .28):\) $$\frac{d \theta}{d t}=\pm \sqrt{\frac{2 g}{L}\left(\cos \theta-\cos \theta_{0}\right)}$$. $$(1.11 .29)$$ (c) Show from Equation \((1.11 .29)\) that the time \(T\) (equal to one-fourth of the period of motion) required for \(\theta\) to go from 0 to \(\theta_{0}\) is given by the elliptic integral of the first kind $$T=\sqrt{\frac{L}{2 g}} \int_{0}^{\theta_{0}} \frac{1}{\sqrt{\cos \theta-\cos \theta_{0}}} d \theta$$. (d) Show that \((1.11 .30)\) can be written as $$T=\sqrt{\frac{L}{g}} \int_{0}^{\pi / 2} \frac{1}{\sqrt{1-k^{2} \sin ^{2} u}} d u$$ $$\begin{aligned} &\text { where } k=\sin \left(\theta_{0} / 2\right) . \text { [Hint: First express } \cos \theta\\\ &\text { and }\left.\cos \theta_{0} \text { in terms of } \sin ^{2}(\theta / 2) \text { and } \sin ^{2}\left(\theta_{0} / 2\right) .\right] \end{aligned}$$

Determine which of the five types of differential equations we have studied the given equation falls into (see Table \(1.12 .1),\) and use an appropriate technique to find the general solution. $$\left(1+2 x e^{y}\right) d x-\left(e^{y}+x\right) d y=0$$

Each spring, sandhill cranes migrate through the Platte River valley in central Nebraska. An estimated maximum of a half-million of these birds reach the region by April 1 each year. If there are only 100,000 sandhill cranes 15 days later and the sandhill cranes leave the Platte River valley at a rate proportional to the number of sandhill cranes still in the valley at the time, (a) How many sandhill cranes remain in the valley 30 days after April \(1 ?\) (b) How many sandhill cranes remain in the valley 35 days after April \(1 ?\) (c) How many days after April 1 will there be less than 1000 sandhill cranes in the valley?

Solve the given differential equation. $$y^{\prime}+2 x^{-1} y=6 \sqrt{1+x^{2}} \sqrt{y}, \quad x>0$$

Determine which of the five types of differential equations we have studied the given equation falls into (see Table \(1.12 .1),\) and use an appropriate technique to find the general solution. $$\frac{d y}{d x}=\frac{\sin y+y \cos x+1}{1-x \cos y-\sin x}$$
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