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Ch. 2 - Limits
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 2 - LimitsProblem 2.32a
Chapter 2, Problem 2.32a

A rock is dropped off the edge of a cliff, and its distance s (in feet) from the top of the cliff after t seconds is s(t)=16t^2. Assume the distance from the top of the cliff to the ground is 96 ft.


a. When will the rock strike the ground? 

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1
We need to find the time \( t \) when the rock hits the ground. This happens when the distance \( s(t) \) equals the height of the cliff, which is 96 feet.
We have the distance function \( s(t) = 16t^2 \). Set this equal to 96 to find when the rock hits the ground: \( 16t^2 = 96 \).
Divide both sides of the equation by 16 to isolate \( t^2 \): \( t^2 = \frac{96}{16} \).
Calculate \( \frac{96}{16} \) to simplify the equation to \( t^2 = 6 \).
Take the square root of both sides to find \( t \): \( t = \sqrt{6} \). Remember to consider only the positive root since time cannot be negative.>

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Quadratic Functions

The function s(t) = 16t^2 is a quadratic function, which represents a parabolic relationship between time t and distance s. In this context, it models the distance fallen by the rock over time due to gravity. Understanding the properties of quadratic functions, such as their vertex and roots, is essential for determining when the rock will hit the ground.
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Roots of Equations

Finding when the rock strikes the ground involves solving for the roots of the equation s(t) = 16t^2. The roots represent the values of t when the distance s equals zero, indicating the moment the rock reaches the ground. This requires setting the equation equal to the height of the cliff (96 ft) and solving for t.
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Physical Interpretation of Motion

In this scenario, the physical interpretation of motion under gravity is crucial. The equation s(t) = 16t^2 derives from the physics of free fall, where the distance fallen is proportional to the square of the time elapsed. Recognizing this relationship helps in understanding the implications of the calculated time when the rock strikes the ground.
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Related Practice
Textbook Question

Complete the following steps for the given functions. 


a. Find the slant asymptote of ff.


f(x)=4x3+4x2+7x+4x2+1f\(\left\)(x\(\right\))=\(\frac{4x^3+4x^2+7x+4}{x^2+1}\)

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Textbook Question

a. Estimate lim x→π/4 cos 2x / cos x − sin x by making a table of values of cos 2x / cos x − sin x for values of x approaching π/4. Round your estimate to four digits.

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Textbook Question

For the following position functions, make a table of average velocities similar to those in Exercises 19–20 and make a conjecture about the instantaneous velocity at the indicated time. 


a. s(t)=−16t^2+80t+60 at t=3

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Textbook Question

A projectile is fired vertically upward and has a position given by s(t)=−16t^2+128t+192, for 0≤t≤9.


a. Graph the position function, for 0≤t≤9.

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Textbook Question

a. Use a graphing utility to estimate lim x→0 tan 2x / sin x, lim x→0 tan 3x / sin x, and lim x→0 tan 4x / sin x.

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Textbook Question

Given the graph of f in the following figures, find the slope of the secant line that passes through (0,0) and (h,f(h))in terms of h, for h>0 and h<0.


f(x)=x1/3 <IMAGE>

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