/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Vertical Motion The height of an... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 14

Vertical Motion The height of an object moving vertically is given by $$s=-16 t^{2}+96 t+112$$ with \(s\) in \(\mathrm{ft}\) and \(t\) in sec. Find (a) the object's velocity when \(t=0,\) (b) its maximum height and when it occurs, and (c) its velocity when \(s=0\) .

Short Answer

Expert verified
The object's velocity at \(t = 0\) is 96 ft/sec. It reaches a maximum height of 160 feet at \(t = 3\) seconds. When the object hits the ground \(s = 0\), its velocity is -128 ft/sec.

Step by step solution

01

Find the object's velocity at \(t = 0\)

The formula for the height of an object thrown vertically up in the air is given by \(s = -16t^2 + 96t + 112\). Here, the velocity of the object is represented by the derivative of the height function. Differentiate the height, \(s(t)\), with respect to time, \(t\), to find the velocity function, \(v(t)\).\n\nThe derivative of the function \(s(t) = -16t^2 + 96t + 112\) with respect to \(t\) is \(v(t) = -32t + 96\). Hence the object's velocity function is \(v(t) = -32t + 96\), where \(v\) is in ft/sec. To find the object's velocity at \(t = 0\), put \(t = 0\) in \(-32t + 96\) which gives \(v = 96\) ft/sec.
02

Calculate the maximum height and when it occurs

To find the maximum height and when it occurs, first set the velocity equation \(-32t + 96 = 0\) and solve for \(t\), after which we substitute this time value in the equation for \(s(t)\) to find the maximum height. Solving the equation for \(t\) gives \(t = 3\) seconds. Substitute \(t = 3\) into the height equation \(s(t) = -16(3)^2 + 96(3) + 112\) to find the maximum height. The maximum height is \(s = 160\) feet. Therefore, the object reaches a maximum height of 160 feet at \(t = 3\) seconds.
03

Find the velocity when \(s = 0\)

To find the velocity when the object hits the ground, \(s = 0\), we first find the time when the object hits the ground by solving the height equation \(-16t^2 + 96t + 112 = 0\) for \(t\). The resulting values of \(t\) are 0 and 7 seconds (ignoring the negative solution). We choose \(t = 7\) seconds because at \(t = 0\), the object was launched. Substituting \(t = 7\) into the velocity equation gives \(v(7) = -32(7) +96 = -128\) ft/sec. So, the object is travelling at a velocity of -128 ft/sec when it hits the ground.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Derivative of a Function
The derivative of a function is a fundamental concept in calculus. Here, it helps us find the rate of change of a position with respect to time, which is called velocity. For vertical motion, the height
  • s(t) = -16t^2 + 96t + 112
represents the position of an object at any time \(t\). To find the velocity, we differentiate this height function with respect to \(t\). The derivative, denoted \(v(t)\), gives us the velocity function:
  • \(v(t) = \frac{d}{dt}(-16t^2 + 96t + 112) = -32t + 96\)
This results in a linear equation for velocity, indicating how fast the object's height changes over time. The velocity equation \(-32t + 96\) tells us the instantaneous speed at any given moment. At time \(t = 0\), substituting this value into the velocity function shows an initial velocity of 96 ft/sec. This initial velocity helps us understand how fast the object is moving upwards right after being launched.
Maximum Height
Finding the maximum height an object reaches is crucial for understanding vertical motion. The maximum height occurs when the velocity is zero because that's the peak point in the upward trajectory before the object starts falling back down. To find this point, we set the velocity function,
  • \(-32t + 96 = 0\)
Setting the equation to zero because the height stops increasing at maximum height helps determine the critical point,
  • \(t = 3\) seconds
This time value is then substituted back into the height equation,
  • \(s(t) = -16(3)^2 + 96(3) + 112\)
Computing this gives us a maximum height of 160 feet. Thus, the object is at its peak height after 3 seconds of upward movement, providing us clear insights into both the speed and duration of that motion upwards.
Velocity at Impact
When an object returns from its peak, it moves back to the ground, impacting it with a certain speed called the velocity at impact. To find this velocity, we need to know when the object's height becomes zero again, which signifies touching the ground. Hence, we solve the height equation for
  • \(-16t^2 + 96t + 112 = 0\)
Finding the roots of this quadratic equation gives the times at which the object is at ground level. Ignoring the initial launch time \(t = 0\), the solution \(t = 7\) seconds indicates the object hits the ground then. Plugging \(t = 7\) into the velocity function informs us:
  • \(v(7) = -32(7) + 96 = -128\) ft/sec
This negative velocity indicates the direction of the impact 鈥� specifically, that the object is moving downward at 128 ft/sec when it hits the ground. This value quantifies how quickly the object was traveling just before impact, critical for understanding the object's terminal pathway.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Moving Shadow A light shines from the top of a pole 50 ft high. A ball is dropped from the same height from a point 30 ft away from the light as shown below. How fast is the ball's shadow moving along the ground 1\(/ 2\) sec later? (Assume the ball falls a distance \(s=16 t^{2}\) in \(t\) sec. $)

Industrial Production (a) Economists often use the expression expression "rate of growth" in relative rather than absolute terms. For example, let \(u=f(t)\) be the number of people in the labor force at time \(t\) in a given industry. (We treat this function as though it were differentiable even though it is an integer-valued step function.) Let \(v=g(t)\) be the average production per person in the labor force at time \(t .\) The total production is then \(y=u v\) . If the labor force is growing at the rate of 4\(\%\) per year year \((d u / d t=\) 0.04\(u\) ) and the production per worker is growing at the rate of 5\(\%\) per year \((d v / d t=0.05 v),\) find the rate of growth of the total production, y. (b) Suppose that the labor force in part (a) is decreasing at the rate of 2\(\%\) per year while the production per person is increasing at the rate of 3\(\%\) per year. Is the total production increasing, or is it decreasing, and at what rate?

Calculus and Geometry How close does the curve \(y=\sqrt{x}\) come to the point \((3 / 2,0) ?[\)Hint: If you minimize the square of the distance, you can avoid square roots.

The Effect of Flight Maneuvers on the Heart The amount of work done by the heart's main pumping chamber, the left ventricle, is given by the equation $$W=P V+\frac{V \delta v^{2}}{2 g}$$ where \(W\) is the work per unit time, \(P\) is the average blood pressure, \(V\) is the volume of blood pumped out during the unit of time, \(\delta("\) delta") is the density of the blood, \(v\) is the average velocity of the exiting blood, and \(g\) is the acceleration of gravity. When \(P, V, \delta,\) and \(v\) remain constant, \(W\) becomes a function of \(g,\) and the equation takes the simplified form $$W=a+\frac{b}{g}(a, b\( constant \))$$ As a member of NASA's medical team, you want to know how sensitive \(W\) is to apparent changes in \(g\) caused by flight maneuvers, and this depends on the initial value of \(g\) . As part of your investigation, you decide to compare the effect on \(W\) of a given change \(d g\) on the moon, where \(g=5.2 \mathrm{ft} / \mathrm{sec}^{2},\) with the effect the same change \(d g\) would have on Earth, where \(g=32\) \(\mathrm{ft} / \mathrm{sec}^{2} .\) Use the simplified equation above to find the ratio of \(d W_{\mathrm{moon}}\) to \(d W_{\mathrm{Earth}}\)

Tin Pest When metallic tin is kept below \(13.2^{\circ} \mathrm{C}\) it slowly becomes brittle and crumbles to a gray powder. Tin objects eventually crumble to this gray powder spontaneously if kept in a cold climate for years. The Europeans who saw tin organ pipes in their churches crumble away years ago called the change tin pest because it seemed to be contagious. And indeed it was, for the gray powder is a catalyst for its own formation. A catalyst for a chemical reaction is a substance that controls the rate of reaction without undergoing any permanent change in itself. An autocatalytic reaction is one whose product is a catalyst for its own formation. Such a reaction may proceed slowly at first if the amount of catalyst present is small and slowly again at the end, when most of the original substance is used up. But in between, when both the substance and its catalyst product are abundant, the reaction proceeds at a faster pace. In some cases it is reasonable to assume that the rate \(v=d x / d t\) of the reaction is proportional both to the amount of the original substance present and to the amount of product. That is, \(v\) may be considered to be a function of \(x\) alone, and $$v=k x(a-x)=k a x-k x^{2}$$ where \(\begin{aligned} x &=\text { the amount of product, } \\ a &=\text { the amount of substance at the beginning, } \\ k &=\text { a positive constant. } \end{aligned}\) At what value of \(x\) does the rate \(v\) have a maximum? What is the maximum value of \(v ?\)
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.