Answer Solution: a. 5 sec Explanation Given: s = s ( t ) = − 16 t 2 − 48 t + 160 Calculation: a. The ball strike the ground when s = s ( t ) = 0 . s = s ( t ) = − 16 t 2 − 48 t + 160 = 0 t 2 − 3 t + 10 = 0 ( t − 5 ) ( t + 2 ) = 0 t = − 2 or t = 5 discard the solution t = − 2 , the strikes the ground after 5 sec . To determine To find: Instantaneous Velocity of a Ball In physics it is shown that the height s of a ball thrown straight down with an initial velocity of 48 ft/sec from rooftop 160 feet high is, s = s ( t ) = − 16 t 2 − 48 t + 160 where t is the elapsed time that the ball is in the air. b. What is the average velocity of the ball from t = 0 to t = 1 ? Answer Solution: b. 64 ft/sec Explanation Given: s = s ( t ) = − 16 t 2 − 48 t + 160 Calculation: b. The average velocity of the ball from t = 0 to t = 1 is, Δs Δt = s ( 1 ) − s ( 0 ) 1 − 0 s ( 1 ) = − 16 − 48 + 160 = 96 s ( 0 ) = − 16 ( 0 ) – 48 ( 0 ) + 160 = 160 Δs Δt = s ( 1 ) − s ( 0 ) 1 − 0 = 96 − 160 1 = − 64 To determine To find: Instantaneous Velocity of a Ball In physics it is shown that the height s of a ball thrown straight down with an initial velocity of 48 ft/sec from rooftop 160 feet high is, s = s ( t ) = − 16 t 2 − 48 t + 160 where t is the elapsed time that the ball is in the air. c. What is the instantaneous velocity of the ball at time t ? Answer Solution: c. − 16 ( 2 t − 3 ) ft/sec Explanation Given: s = s ( t ) = − 16 t 2 − 48 t + 160 Calculation: c. The instantaneous velocity of the ball at time t 1 is the derivative s ’ ( t 1 ) : that is, s ’ ( t 1 ) = lim t → t 1 s ( t ) − s ( t 1 ) t − t 1 = lim t → t 1 ( − 16 t 2 − 48 t + 160 ) − ( − 16 t 1 2 − 48 t 1 + 160 ) t − t 1 = lim t → t 1 − 16 ( t 2 − t 1 2 − 3 t − 3 t 1 ) t − t 1 = lim t → t 1 − 16 [ ( t + t 1 ) ( t − t 1 ) − 3 ( t − t 1 ) ] t − t 1 = lim t → t 1 − 16 [ ( t + t 1 − 3 ) ( t − t 1 ) ] t − t 1 = lim t → t 1 − 16 ( t + t 1 − 3 ) = − 16 ( 2 t 1 – 3 ) ft/sec Replace t 1 t . the instantaneous velocity of the ball at time t is, s ’ ( t ) = − 16 ( 2 t – 3 ) ft/sec To determine To find: Instantaneous Velocity of a Ball In physics it is shown that the height s of a ball thrown straight down with an initial velocity of 48 ft/sec from rooftop 160 feet high is, s = s ( t ) = − 16 t 2 − 48 t + 160 where t is the elapsed time that the ball is in the air. d. What is the instantaneous velocity of the ball at t = 1 ? Answer Solution: d. 16 ft/sec Explanation Given: s = s ( t ) = − 16 t 2 − 48 t + 160 Calculation: d. The instantaneous velocity of the ball at t = 1 is, s ’ ( t ) = − 16 ( 2 ( 1 ) – 3 ) = − 16 ( − 1 ) = 16 ft/sec To determine To find: Instantaneous Velocity of a Ball In physics it is shown that the height s of a ball thrown straight down with an initial velocity of 48 ft/sec from rooftop 160 feet high is, s = s ( t ) = − 16 t 2 − 48 t + 160 where t is the elapsed time that the ball is in the air. e. What is the instantaneous velocity of the ball when it strikes the ground? Answer Solution: e. − 48 ft/sec Explanation Given: s = s ( t ) = − 16 t 2 − 48 t + 160 Calculation: e. The ball strikes the ground when t = 3 , the instantaneous velocity when t = 3 is, s ’ ( 3 ) = − 16 ( 2 ( 3 ) – 3 ) = − 48 ft/sec The velocity of the ball at t = 3 sec is – 48 ft/sec , the negative value implies that the ball is travelling downward.

9th Edition

ISBN: 9780321716835

Author: Michael Sullivan

Publisher: Addison Wesley

See similar textbooks

Videos

Question

Chapter 14.4, Problem 48AYU

To determine

To find: Instantaneous Velocity of a Ball In physics it is shown that the height $s$ of a ball thrown straight down with an initial velocity of $48 ft/sec$ from rooftop 160 feet high is,

$s = s (t) = - 16 t^{2} - 48 t + 160$

where $t$ is the elapsed time that the ball is in the air.

a. When does the ball strike the ground? That is, how long is the ball in the air?

Expert Solution

Answer to Problem 48AYU

Solution:

a. $5 sec$

Explanation of Solution

Given:

$s = s (t) = - 16 t^{2} - 48 t + 160$

Calculation:

a. The ball strike the ground when $s = s (t) = 0$ .

$s = s (t) = - 16 t^{2} - 48 t + 160 = 0$

$t^{2} - 3 t + 10 = 0$

$(t - 5) (t + 2) = 0$

$t = - 2 or t = 5$

discard the solution $t = - 2$ , the strikes the ground after $5 sec$ .

To determine

To find: Instantaneous Velocity of a Ball In physics it is shown that the height $s$ of a ball thrown straight down with an initial velocity of $48 ft/sec$ from rooftop 160 feet high is,

$s = s (t) = - 16 t^{2} - 48 t + 160$

where $t$ is the elapsed time that the ball is in the air.

b. What is the average velocity of the ball from $t = 0 to t = 1$ ?

Expert Solution

Answer to Problem 48AYU

Solution:

b. $64 ft/sec$

Explanation of Solution

Given:

$s = s (t) = - 16 t^{2} - 48 t + 160$

Calculation:

b. The average velocity of the ball from $t = 0 to t = 1$ is,

$\frac{Δs}{Δt} = \frac{s (1) - s (0)}{1 - 0}$

$s (1) = - 16 - 48 + 160 = 96$

$s (0) = - 16 (0) - 48 (0) + 160 = 160$

$\frac{Δs}{Δt} = \frac{s (1) - s (0)}{1 - 0} = \frac{96 - 160}{1} = - 64$

To determine

To find: Instantaneous Velocity of a Ball In physics it is shown that the height $s$ of a ball thrown straight down with an initial velocity of $48 ft/sec$ from rooftop 160 feet high is,

$s = s (t) = - 16 t^{2} - 48 t + 160$

where $t$ is the elapsed time that the ball is in the air.

c. What is the instantaneous velocity of the ball at time $t$ ?

Expert Solution

Answer to Problem 48AYU

Solution:

c. $- 16 (2 t - 3) ft/sec$

Explanation of Solution

Given:

$s = s (t) = - 16 t^{2} - 48 t + 160$

Calculation:

c. The instantaneous velocity of the ball at time $t 1$ is the derivative $s ’ (t_{1})$ : that is,

$s ’ (t_{1}) = \lim_{t \to t_{1}} \frac{s (t) - s (t_{1})}{t - t_{1}}$

$= \lim_{t \to t_{1}} \frac{(- 16 t^{2} - 48 t + 160) - (- 16 t_{1}^{2} - 48 t_{1} + 160)}{t - t_{1}}$

$= \lim_{t \to t_{1}} \frac{- 16 (t^{2} - t_{1}^{2} - 3 t - 3 t_{1})}{t - t_{1}}$

$= \lim_{t \to t_{1}} \frac{- 16 [(t + t_{1}) (t - t_{1}) - 3 (t - t_{1})]}{t - t_{1}}$

$= \lim_{t \to t_{1}} \frac{- 16 [(t + t_{1} - 3) (t - t_{1})]}{t - t_{1}}$

$= \lim_{t \to t_{1}} - 16 (t + t_{1} - 3)$

$= - 16 (2 t_{1} - 3) ft/sec$

Replace $t_{1} t$ . the instantaneous velocity of the ball at time t is,

$s ’ (t) = - 16 (2 t - 3) ft/sec$

To determine

To find: Instantaneous Velocity of a Ball In physics it is shown that the height $s$ of a ball thrown straight down with an initial velocity of $48 ft/sec$ from rooftop 160 feet high is,

$s = s (t) = - 16 t^{2} - 48 t + 160$

where $t$ is the elapsed time that the ball is in the air.

d. What is the instantaneous velocity of the ball at $t = 1$ ?

Expert Solution

Answer to Problem 48AYU

Solution:

d. $16 ft/sec$

Explanation of Solution

Given:

$s = s (t) = - 16 t^{2} - 48 t + 160$

Calculation:

d. The instantaneous velocity of the ball at $t = 1$ is,

$s ’ (t) = - 16 (2 (1) - 3) = - 16 (- 1) = 16 ft/sec$

To determine

To find: Instantaneous Velocity of a Ball In physics it is shown that the height $s$ of a ball thrown straight down with an initial velocity of $48 ft/sec$ from rooftop 160 feet high is,

$s = s (t) = - 16 t^{2} - 48 t + 160$

where $t$ is the elapsed time that the ball is in the air.

e. What is the instantaneous velocity of the ball when it strikes the ground?

Expert Solution

Answer to Problem 48AYU

Solution:

e. $- 48 ft/sec$

Explanation of Solution

Given:

$s = s (t) = - 16 t^{2} - 48 t + 160$

Calculation:

e. The ball strikes the ground when $t = 3$ , the instantaneous velocity when $t = 3$ is,

$s ’ (3) = - 16 (2 (3) - 3) = - 48 ft/sec$

The velocity of the ball at $t = 3 sec$ is $- 48 ft/sec$ , the negative value implies that the ball is travelling downward.

Chapter 14.4, Problem 47AYU

Chapter 14.4, Problem 49AYU

Chapter 14 Solutions

Precalculus

Ch. 14.1 - Graph f( x )={ 3x2ifx2 3ifx=2 (pp.100-102)Ch. 14.1 - Prob. 2AYU Ch. 14.1 - Prob. 3AYU Ch. 14.1 - Prob. 4AYU Ch. 14.1 - True or False lim xc f( x )=N may be described by...Ch. 14.1 - Prob. 6AYU Ch. 14.1 - lim x2 ( 4 x 3 )Ch. 14.1 - lim x3 ( 2 x 2 +1 )Ch. 14.1 - lim x0 x+1 x 2 +1 Ch. 14.1 - Prob. 10AYU

Additional Math Textbook Solutions

Find more solutions based on key concepts

The height of the square pyramid whose volume and length are 873.18 m3,12.6 m respectively.

Pre-Algebra Student Edition

CHECK POINT I You deposit $1000 in a saving account at a bank that has a rate of 4%. a. Find the amount, A, of ...

Thinking Mathematically (6th Edition)

A retail establishment accepts either the American Express or the VISA credit card. A total of 24 percent of it...

A First Course in Probability (10th Edition)

Explain the meaning of the term “statistically significant difference” in statistics terminology.

Intro Stats, Books a la Carte Edition (5th Edition)

The following set of data is from sample of n=5: a. Compute the mean, median, and mode. b. Compute the range, v...

Basic Business Statistics, Student Value Edition

1. combination of numbers, variables, and operation symbols is called an algebraic______.

Algebra and Trigonometry (6th Edition)

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.