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Chapter 8: Problem 30

An object is fired vertically upward with an initial velocity \(v(0)=v_{0}\) from an initial position \(s(0)=s_{0}\). a. For the following values of \(v_{0}\) and \(s_{0},\) find the position and velocity functions for all times at which the object is above the ground. b. Find the time at which the highest point of the trajectory is reached and the height of the object at that time. $$v_{0}=49 \mathrm{m} / \mathrm{s}, s_{0}=60 \mathrm{m}$$

Short Answer

Expert verified
Answer: The position and velocity functions are given by \(s(t) = 60 + 49t - 4.9t^2\) and \(v(t) = 49 - 9.8t\). The object reaches its highest point at approximately \(t = 5\mathrm{s}\) and the height at that time is approximately \(180.5\mathrm{m}\).

Step by step solution

01

1. Recall the position and velocity equations for an object in free fall

The position and velocity of an object in free fall are given by: $$s(t) = s_0 + v_0t - \frac{1}{2}gt^2$$ $$v(t) = v_0 - gt$$ where \(s_0\) is the initial position, \(v_0\) is the initial velocity, \(g\) is the acceleration due to gravity (approximately \(9.8 \mathrm{m/s^2}\)), and \(t\) is time.
02

2. Find the position and velocity functions using the given initial values

We are provided with the initial values \(v_0 = 49\mathrm{m/s}\) and \(s_0 = 60\mathrm{m}\). Using those, we can define the position and velocity functions for \(t \geq 0\): $$s(t) = 60 + 49t - 4.9t^2$$ $$v(t) = 49 - 9.8t$$
03

3. Find the time when the object reaches its highest point

At the highest point, the object's vertical velocity will be zero. Therefore, we can solve the equation \(v(t) = 0\) for \(t\): $$0 = 49 - 9.8t$$ $$t = \frac{49}{9.8}$$ \(t \approx 5\mathrm{s}\)
04

4. Find the height of the object at the highest point

Using the position function \(s(t)\) and the time at the highest point, we can find the height of the trajectory's highest point: $$s(5) = 60 + 49(5) - 4.9(5)^2$$ \(s(5) \approx 180.5\mathrm{m}\) #step 1.# The position and velocity functions for all times the object is above the ground are: $$s(t) = 60 + 49t - 4.9t^2$$ $$v(t) = 49 - 9.8t$$ #step 2.# The time at which the highest point is reached is about \(t = 5\mathrm{s}\), and at that time, the object is about \(180.5\mathrm{m}\) above the ground.

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Key Concepts

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

Kinematic Equations
Kinematic equations are a set of formulas that connect the five key parameters of motion: displacement, initial velocity, final velocity, acceleration, and time. These equations are particularly useful for analyzing one-dimensional motion with constant acceleration. In the context of projectile motion, they help us describe the behavior of objects in free fall or those launched into the air.

For vertical motion, the two key kinematic equations we often use relate to position and velocity:
  • Position: \( s(t) = s_0 + v_0t - \frac{1}{2}gt^2 \)
  • Velocity: \( v(t) = v_0 - gt \)

Here, \( s(t) \) and \( v(t) \) represent the position and velocity of the object at time \( t \). \( s_0 \) and \( v_0 \) are the initial position and velocity, respectively. The term \( g \) denotes the acceleration due to gravity, typically around \( 9.8 \mathrm{m/s^2} \). These equations allow us to predict where and how fast the object will be at any given time.
Initial Velocity
Initial velocity is the speed at which an object starts its motion. It's a crucial factor in determining how far and how high an object will travel in projectile motion.

In the exercise, the initial velocity \( v_0 \) is given as \( 49 \, \mathrm{m/s} \). This initial push upward impacts both how quickly the object rises and how long it stays in the air before gravity pulls it back down to the ground.
  • If the initial velocity is greater, the object will reach a higher point and will stay airborne longer.
  • If it is lower, the object will achieve a lower height and will be in the air for a shorter duration.

Understanding initial velocity helps us calculate where the object will be at any moment using kinematic equations.
Acceleration due to Gravity
The acceleration due to gravity \(g\) is the rate at which an object's velocity changes when it is in free fall. It is a constant that pulls objects downward towards the Earth's surface at approximately \(9.8 \mathrm{m/s^2}\). This force is the same for all objects, regardless of their mass.

When dealing with projectile motion, this acceleration is considered negative since it acts in the opposite direction to the initial velocity, which is upward. Over time, gravity reduces the object's upward speed, brings it to a temporary stop at the highest point, and then increases its speed again as it falls back down.

With the use of kinematic equations, incorporating gravity allows us to model the object's deceleration on the way up and its acceleration on the way down.
Free Fall
Free fall describes the motion of an object under the influence of gravitational force only. When something is in free fall, it's either dropped or released without any initial force except for gravity acting upon it.

In input exercises like projectile problems, objects often have an initial upward velocity, as is the case here. However, gravity acts to decelerate the object immediately until it reaches the highest point of its path where its velocity is zero.
  • At this point, the object starts to accelerate downwards towards the ground.
  • Free fall takes into account only gravitational acceleration, hence no air resistance or other forces are factored in.

This idealized motion is key to understanding the symmetrical ascent and descent of projectiles in physics.

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Most popular questions from this chapter

Analytical solution of the predator-prey equations The solution of the predator-prey equations $$x^{\prime}(t)=-a x+b x y, y^{\prime}(t)=c y-d x y$$ can be viewed as parametric equations that describe the solution curves. Assume that \(a, b, c,\) and \(d\) are positive constants and consider solutions in the first quadrant. a. Recalling that \(\frac{d y}{d x}=\frac{y^{\prime}(t)}{x^{\prime}(t)},\) divide the first equation by the second equation to obtain a separable differential equation in terms of \(x\) and \(y\) b. Show that the general solution can be written in the implicit form \(e^{d x+b y}=C x^{c} y^{a},\) where \(C\) is an arbitrary constant. c. Let \(a=0.8, b=0.4, c=0.9,\) and \(d=0.3 .\) Plot the solution curves for \(C=1.5,2,\) and \(2.5,\) and confirm that they are, in fact, closed curves. Use the graphing window \([0,9] \times[0,9]\)

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Let \(y(t)\) be the concentration of a substance in a chemical reaction (typical units are moles/liter). The change in the concentration, under appropriate conditions, is modeled by the equation \(\frac{d y}{d t}=-k y^{n},\) where \(k>0\) is a rate constant and the positive integer \(n\) is the order of the reaction. a. Show that for a first-order reaction \((n=1)\), the concentration obeys an exponential decay law. b. Solve the initial value problem for a second-order reaction \((n=2)\) assuming \(y(0)=y_{0}\) c. Graph the concentration for a first-order and second-order reaction with \(k=0.1\) and \(y_{0}=1\)

A physiological model A common assumption in modeling drug assimilation is that the blood volume in a person is a single compartment that behaves like a stirred tank. Suppose that the blood volume is a four-liter tank that initially has a zero concentration of a particular drug. At time \(t=0,\) an intravenous line is inserted into a vein (into the tank) that carries a drug solution with a concentration of \(500 \mathrm{mg} / \mathrm{L} .\) The inflow rate is \(0.06 \mathrm{L} / \mathrm{min}\). Assume that the drug is quickly mixed thoroughly in the blood and that the volume of blood remains constant. a. Write an initial value problem that models the mass of the drug in the blood, for \(t \geq 0\) b. Solve the initial value problem and graph both the mass of the drug and the concentration of the drug. c. What is the steady-state mass of the drug in the blood? d. After how many minutes does the drug mass reach \(90 \%\) of its steady-state level?

The following models were discussed in Section 1 and reappear in later sections of this chapter. In each case carry out the indicated analysis using direction fields. A model that describes the free fall of an object in a gravitational field subject to air resistance uses the equation \(v^{\prime}(t)=g-b v,\) where \(v(t)\) is the velocity of the object, for \(t \geq 0, g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, and \(b>0\) is a constant that involves the mass of the object and the air resistance. Let \(b=0.1 \mathrm{s}^{-1}\). a. Draw the direction field for \(0 \leq t \leq 60,0 \leq y \leq 150\). b. For what initial values \(v(0)=A\) are solutions increasing? Decreasing? c. What is the equilibrium solution?
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