Skip to main content
Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 4.1.61
Chapter 4, Problem 4.1.61

Theory and Examples


Maximum height of a vertically moving body The height of a body moving vertically is given by s = −12gt² + υ₀t + s₀,  g > 0, with s in meters and t in seconds. Find the body’s maximum height.

Verified step by step guidance
1
Identify the given equation for the height of the body: \( s(t) = -12gt^2 + v_0t + s_0 \), where \( g \) is the acceleration due to gravity, \( v_0 \) is the initial velocity, and \( s_0 \) is the initial height.
To find the maximum height, we need to determine the time \( t \) at which the velocity is zero. The velocity \( v(t) \) is the derivative of the height function \( s(t) \) with respect to time \( t \).
Differentiate the height function: \( v(t) = \frac{d}{dt}(-12gt^2 + v_0t + s_0) = -24gt + v_0 \).
Set the velocity \( v(t) \) to zero to find the critical point: \( -24gt + v_0 = 0 \). Solve for \( t \) to find \( t = \frac{v_0}{24g} \).
Substitute \( t = \frac{v_0}{24g} \) back into the original height equation \( s(t) \) to find the maximum height: \( s_{max} = -12g\left(\frac{v_0}{24g}\right)^2 + v_0\left(\frac{v_0}{24g}\right) + s_0 \). Simplify this expression to find the maximum height.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4m
Was this helpful?

Key Concepts

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

Quadratic Functions

A quadratic function is a polynomial of degree two, typically in the form of ax² + bx + c. The graph of a quadratic function is a parabola, which can open upwards or downwards. In the context of the problem, the height function s = −12gt² + υ₀t + s₀ is a quadratic function of time t, where the coefficient of t² determines the direction of the parabola.
Recommended video:
6:04
Introduction to Polynomial Functions

Vertex of a Parabola

The vertex of a parabola is the point where it reaches its maximum or minimum value. For a quadratic function in the form ax² + bx + c, the vertex can be found using the formula t = -b/(2a). In this problem, the vertex represents the time at which the body reaches its maximum height, as the parabola opens downwards due to the negative coefficient of t².
Recommended video:
7:42
Properties of Parabolas

Derivative and Critical Points

The derivative of a function gives the rate of change and can be used to find critical points where the function's slope is zero. These points are potential maxima or minima. For the height function, taking the derivative with respect to time and setting it to zero helps find the time at which the maximum height occurs, confirming the vertex calculation.
Recommended video:
04:50
Critical Points
Related Practice
Textbook Question

Finding Indefinite Integrals


In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.


∫(cscθ cotθ) / 2 dθ

20
views
Textbook Question

[Technology Exercises] When solving Exercises 14–30, you may need to use appropriate technology (such as a calculator or a computer).

26. Factoring a quartic Find the approximate values of r_1 through r_4 in the factorization

8x^4-14x^3-9x^2+11x-1=8(x-r_1)(x-r_2)(x-r_3)(x-r_4)

207
views
Textbook Question

Business and Economics

62. Production level Suppose that c(x)=x^3-20x^2 + 20,000x is the cost of manufacturing x items. Find a production level that will minimize the average cost of making x items.

182
views
Textbook Question

The 8-ft wall shown here stands 27 ft from the building. Find the length of the shortest straight beam that will reach to the side of the building from the ground outside the wall.

211
views
Textbook Question

Finding Indefinite Integrals


In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.


∫(−3csc²x)dx

17
views
Textbook Question

Initial Value Problems

Solve the initial value problems in Exercises 71–90.

d³y/dx³ = 6; y″(0) = −8, y′(0) = 0, y(0) = 5

14
views