Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals3h 25m
4. Applications of Derivatives
Motion Analysis
Problem 3.6.24c
Textbook Question
Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 19.6 m/s from a height of 24.5 m above the ground. The height (in meters) of the stone above the ground t seconds after it is thrown is s(t) = -4.9t²+19.6t+24.5.
c. What is the height of the stone at the highest point?
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1
Identify the function that describes the height of the stone, which is given as s(t) = -4.9t² + 19.6t + 24.5.
Recognize that the height of the stone reaches its maximum when the velocity is zero. To find this, first calculate the derivative of s(t) with respect to t, which gives you s'(t).
Set the derivative s'(t) equal to zero to find the time t at which the stone reaches its highest point. This will involve solving the equation -9.8t + 19.6 = 0.
Once you find the value of t, substitute it back into the original height function s(t) to find the maximum height of the stone.
Evaluate s(t) at the calculated time to determine the height of the stone at its highest point.
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