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 $48\text{ ft/sec}$ from rooftop 160 feet high is,

$s=s\left(t\right)=-16t^2-48t+160$

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?

Answer to Problem 48AYU

Solution:

a. $5\text{sec}$

Explanation of Solution

Given:

$s=s\left(t\right)=-16t^2-48t+160$

Calculation:

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

$s=s\left(t\right)=-16t^2-48t+160=0$

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

$\left(t-5\right)\left(t+2\right)=0$

$t=-2\text{or}t=5$

discard the solution $t=-2$, the strikes the ground after $5\text{sec}$.

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 $48\text{ ft/sec}$ from rooftop 160 feet high is,

$s=s\left(t\right)=-16t^2-48t+160$

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

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

Answer to Problem 48AYU

Solution:

b. $64\text{ft/sec}$

Explanation of Solution

Given:

$s=s\left(t\right)=-16t^2-48t+160$

Calculation:

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

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

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

$s\left(0\right)=-16\left(0\right)–48\left(0\right)+160=160$

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

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 $48\text{ ft/sec}$ from rooftop 160 feet high is,

$s=s\left(t\right)=-16t^2-48t+160$

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 to Problem 48AYU

Solution:

c. $-16\left(2t-3\right)\text{ft/sec}$

Explanation of Solution

Given:

$s=s\left(t\right)=-16t^2-48t+160$

Calculation:

c. The instantaneous velocity of the ball at time $$t1$$ is the derivative $s’\left(t_1\right)$: that is,

$s’\left(t_1\right)=\underset{t\to t_1}{\lim }\frac{s\left(t\right)-s\left(t_1\right)}{t-t_1}$

$=\underset{t\to t_1}{\lim }\frac{\left(-16t^2-48t+160\right)-\left(-16t_1^2-48t_1+160\right)}{t-t_1}$

$=\underset{t\to t_1}{\lim }\frac{-16\left(t^2-t_1^2-3t-3t_1\right)}{t-t_1}$

$=\underset{t\to t_1}{\lim }\frac{-16\left[\left(t+t_1\right) \left(t-t_1\right)-3 \left(t-t_1\right)\right]}{t-t_1}$

$=\underset{t\to t_1}{\lim }\frac{-16\left[\left(t+t_1-3\right) \left(t-t_1\right)\right]}{t-t_1}$

$=\underset{t\to t_1}{\lim }-16\left(t+t_1-3\right)$

$=-16\left(2t_1–3\right)\text{ft/sec}$

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

$s’\left(t\right)=-16\left(2t–3\right)\text{ ft/sec}$

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 $48\text{ ft/sec}$ from rooftop 160 feet high is,

$s=s\left(t\right)=-16t^2-48t+160$

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 to Problem 48AYU

Solution:

d. $16\text{ft/sec}$

Explanation of Solution

Given:

$s=s\left(t\right)=-16t^2-48t+160$

Calculation:

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

$s’\left(t\right)=-16\left(2\left(1\right)–3\right)=-16\left(-1\right)=16\text{ ft/sec}$

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 $48\text{ ft/sec}$ from rooftop 160 feet high is,

$s=s\left(t\right)=-16t^2-48t+160$

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 to Problem 48AYU

Solution:

e. $-48\text{ft/sec}$

Explanation of Solution

Given:

$s=s\left(t\right)=-16t^2-48t+160$

Calculation:

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

$s’\left(3\right)=-16\left(2\left(3\right)–3\right)=-48\text{ ft/sec}$

The velocity of the ball at $t=3\text{sec}$ is $–48\text{ft/sec}$, the negative value implies that the ball is travelling downward.