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Chapter 3: Problem 29

Object dropped from a tower An object is dropped from the top of a \(100-\) m-high tower. Its height above ground after \(t\) sec is \(100-4.9 t^{2}\) m. How fast is it falling 2 sec after it is dropped?

Short Answer

Expert verified
The object is falling at 19.6 m/s after 2 seconds.

Step by step solution

01

Understand the problem

We have an object that is dropped from a height of 100 meters. We are given an equation for the height of the object above the ground as a function of time, specifically, \[h(t) = 100 - 4.9 t^2.\]We need to find the rate at which the object is falling (the velocity) at \(t = 2\) seconds.
02

Differentiate the height equation

The rate of change of the object's height with respect to time gives us its velocity. We find this by differentiating the height equation:\[h(t) = 100 - 4.9 t^2.\]Differentiating this with respect to \(t\) gives:\[v(t) = \frac{d}{dt}(100 - 4.9t^2) = -9.8t.\]
03

Substitute t=2 into the velocity equation

Now that we have the velocity function, \(v(t) = -9.8t\), we substitute \(t = 2\) to find the velocity at this time:\[v(2) = -9.8 \times 2 = -19.6.\]This means the object is falling at a rate of 19.6 meters per second after 2 seconds.

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus that involves calculating the rate at which a function's output changes with respect to changes in its input. In simple terms, differentiation allows us to find the slope of a function at any point, which can help analyze how things change over time or space.
In the context of physics and motion, differentiation is crucial for finding quantities like velocity and acceleration from position or height functions. For our exercise, we have the position of the object, given by the function \[ h(t) = 100 - 4.9 t^2. \]
To find the object's velocity, we need the rate of change of height with respect to time. This is done through differentiation, giving us the velocity function:\[ v(t) = \frac{d}{dt}(100 - 4.9 t^2) = -9.8t. \]
Here, \(-9.8t\) represents how fast the object鈥檚 height is decreasing over time, providing us a clear insight into the object's falling speed at any time \(t\).
Velocity
Velocity is an important concept in both calculus and physics. It measures how fast an object changes its position over time in a specific direction and is therefore a vector quantity. Unlike speed, which is scalar and only tells us how fast an object is moving, velocity also provides information about the direction of movement.
In our problem, the velocity function derived from the height function is:\[ v(t) = -9.8t. \]
By substituting specific values of \(t\), we can find the velocity at any given moment. For \(t = 2\) seconds, the computation \[ v(2) = -9.8 \times 2 = -19.6 \] tells us that the object's velocity is \(-19.6\) meters per second. The negative sign indicates that the object is moving towards the ground, or downwards.
The velocity value helps us understand the dynamics of the object's fall at particular moments, making it an indispensable tool for studying motion.
Motion in Physics
Motion in physics refers to the change in position of an object over time. It can be described in terms of displacement, velocity, and acceleration, involving various principles to understand how objects move.
In our exercise, we have a scenario where an object is dropped from a tower, and its motion is described using the height function: \[ h(t) = 100 - 4.9t^2. \]
This function captures the essence of motion under gravity, with a constant acceleration of \(9.8 \, \text{m/s}^2\), as seen by the coefficient \(4.9\) (half of 9.8) in the equation. As time increases, the term \(4.9t^2\) grows, representing increased distance fallen due to gravity's pull.
When studying motion, it is important to consider concepts like initial position, velocity (as derived), and acceleration, as these components reveal the path and speed of moving objects. The results from the differentiation of the height function show how the object continuously accelerates towards the ground, modeled accurately by physics equations.

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Most popular questions from this chapter

In Exercises \(35-40,\) write a differential formula that estimates the given change in volume or surface area. $$ \begin{array}{l}{\text { The change in the lateral surface area } S=\pi r \sqrt{r^{2}+h^{2}} \text { of a right }} \\ {\text { circular cone when the radius changes from } r_{0} \text { to } r_{0}+d r \text { and the }} \\\ {\text { height does not change }}\end{array} $$

Use a CAS to perform the following steps for the functions \begin{equation} \begin{array}{l}{\text { a. Plot } y=f(x) \text { to see that function's global behavior. }} \\ {\text { b. Define the difference quotient } q \text { at a general point } x, \text { with }} \\ {\text { general step size } h .} \\\ {\text { c. Take the limit as } h \rightarrow 0 . \text { What formula does this give? }} \\ {\text { d. Substitute the value } x=x_{0} \text { and plot the function } y=f(x)} \\ \quad {\text { together with its tangent line at that point. }} \\ {\text { e. Substitute various values for } x \text { larger and smaller than } x_{0} \text { into }} \\ \quad {\text { the formula obtained in part (c). Do the numbers make sense }} \\ \quad {\text { with your picture? }} \\ {\text { f. Graph the formula obtained in part (c). What does it mean }} \\ \quad {\text { when its values are negative? Zero? Positive? Does this make }} \\ \quad {\text { sense with your plot from part (a)? Give reasons for your }} \\ \quad {\text { answer. }}\end{array} \end{equation} $$f(x)=\sin 2 x, \quad x_{0}=\pi / 2$$

Chain Rule Suppose that \(f(x)=x^{2}\) and \(g(x)=|x| .\) Then the composites $$(f \circ g)(x)=|x|^{2}=x^{2} \quad\( and \)\quad(g \circ f)(x)=\left|x^{2}\right|=x^{2}$$ are both differentiable at \(x=0\) even though \(g\) itself is not differentiable at \(x=0 .\) Does this contradict the Chain Rule? Explain.

Motion in the plane The coordinates of a particle in the metric \(x y\)-plane are differentiable functions of time \(t\) with \(d x / d t=\) \(-1 \mathrm{m} / \mathrm{sec}\) and \(d y / d t=-5 \mathrm{m} / \mathrm{sec} .\) How fast is the particle's distance from the origin changing as it passes through the point \((5,12) ?\)

Assume that \(2 x+3 y=12\) and \(d y / d t=-2 .\) Find \(d x / d t.\)
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