# Question bab32

Nov 15, 2014

When an object is dropped so that only the force of gravity is acting on it, the symbol for acceleration becomes $g$, and is ${\text{-9.81m/s}}^{2}$

Example 1. Dropping a book from a height of $\text{1.50}$ meters.
How long will it take the book to land, and what will its final velocity be?

Known/Given:
$g$ = ${\text{-9.81m/s}}^{2}$
$d$ = $\text{-1.50m}$
v"_i = $\text{0m/s}$

Unknown:
time, $t$
final velocity v"_f

Equations:
${\text{v"_f}}^{2}$ = ${\text{v"_i}}^{2}$ + $\text{2} g d$
${\text{v}}_{f}$ = ${\text{v}}_{i}$ + $g$$t$

Solution:
First, determine the final velocity of the book when it lands using the kinematic equation ${\text{v"_f}}^{2}$ = ${\text{v"_i}}^{2}$ + $\text{2} g d$, and solving for v"_f. Since the initial velocity is zero, the equation becomes

${\text{v"_f}}^{2}$ = $\text{2} g d$

v"_f = sqrt$\text{2} g d$ = sqrt$\left(\text{2}\right)$$\left({\text{-9.81m/s}}^{2}\right)$$\left(\text{-1.50m}\right)$ = sqrt${\text{29.43m"^2"/s}}^{2}$ = $\text{-5.42m/s}$
(Remember that a square root is + or -, and the downward motion is negative.)

Now determine the time it takes for the book to land using the equation ${\text{v}}_{f}$ = ${\text{v}}_{i}$ + $g$$t$ and solving for $t$.
Since v"_i is 0m/s, the equation becomes v"_f = $g$$t$.

$t$ = ${\text{v}}_{f} / g$ = ${\text{-5.42m/s"/"-9.81m/s}}^{2}$ = 0.552s"

So, if you drop a book from a height of $\text{1.50m}$, such that only gravity is acting on it, the time it takes to land is $\text{0.552s}$, and its final velocity is $\text{-5.42m/s}$.

Example 2. Throwing a book down from a height of $\text{1.50m}$.
Lets say you threw a book down with an initial velocity of $\text{-2.00m/s}$ and an acceleration of ${\text{-10.50m/s}}^{2}$.
How long will it take the book to land, and what will its final velocity be?

Known/Given:
$a$ = ${\text{-10.50m/s}}^{2}$
$d$ = $\text{-1.50m}$
v"_i = $\text{-2.00m/s}$

Unknown:
time, $t$
final velocity v"_f

Equations:
${\text{v"_f}}^{2}$ = ${\text{v"_i}}^{2}$ + $\text{2} a d$
${\text{v}}_{f}$ = ${\text{v}}_{i}$ + $a t$

Solution:

First, determine the final velocity of the book when it lands using the kinematic equation ${\text{v"_f}}^{2}$ = ${\text{v"_i}}^{2}$ + $\text{2} a d$, and solving for v"_f.

${\text{v"_f}}^{2}$ = (${\text{-2.00m/s)}}^{2}$ + "("2")$\left({\text{-10.50m/s}}^{2}\right)$$\left(\text{-1.50m}\right)$ = ${\text{35.50m"^2"/s}}^{2}$

$\text{v"_f}$ = sqrt${\text{35.50m"^2"/s}}^{2}$ = $\text{-5.96m/s}$
(Again, remember that a square root is + or -, and here the downward motion is negative.)

Now determine the time it takes the thrown book to land using the equation ${\text{v}}_{f}$ = ${\text{v}}_{i}$ + $a t$, and solving for $t$.

$t$ = $\frac{{\text{v"_f - "v}}_{i}}{a}$ = $\text{-5.96m/s - (-2.00m/s)"/"-10.50m/s"^2}$ = $\text{0.337s}$

So a book thrown down from a height of $\text{1.50m}$ with an initial velocity of $\text{-2.00m/s}$ and an acceleration of $\text{-10.50m/s"^2}$, will land after 0.337s" with a final velocity of $\text{-5.96m/s}$. This is a shorter time and a faster final velocity than when the book was dropped.

Nov 15, 2014

Yes. When an object is dropped, its initial velocity is zero and its acceleration is that of gravity, which is ${\text{-9.81m/s}}^{2}$. When an object is thrown to the ground, its initial velocity is greater than zero and its acceleration is greater than ${\text{-9.81m/s}}^{2}$.

There are four kinematic equations that are used to study motion in one direction. .

Two of these equations will be used to test how fast an object falls, and its final velocity.

Note: When an object is dropped so that only the force of gravity is acting on it, the symbol for acceleration becomes $g$, and is ${\text{-9.81m/s}}^{2}$

When an object is dropped so that only the force of gravity is acting on it, the symbol for acceleration becomes $g$, and is ${\text{-9.81m/s}}^{2}$

Example 1. Dropping a book from a height of $\text{1.50}$ meters.
How long will it take the book to land, and what will its final velocity be?

Known/Given:
$g$ = ${\text{-9.81m/s}}^{2}$
$d$ = $\text{-1.50m}$
v"_i = $\text{0m/s}$

Unknown:
time, $t$
final velocity v"_f

Equations:
${\text{v"_f}}^{2}$ = ${\text{v"_i}}^{2}$ + $\text{2} g d$
${\text{v}}_{f}$ = ${\text{v}}_{i}$ + $g$$t$

Solution:
First, determine the final velocity of the book when it lands using the kinematic equation ${\text{v"_f}}^{2}$ = ${\text{v"_i}}^{2}$ + $\text{2} g d$, and solving for v"_f. Since the initial velocity is zero, the equation becomes

${\text{v"_f}}^{2}$ = $\text{2} g d$

v"_f = sqrt$\text{2} g d$ = sqrt$\left(\text{2}\right)$$\left({\text{-9.81m/s}}^{2}\right)$$\left(\text{-1.50m}\right)$ = sqrt${\text{29.43m"^2"/s}}^{2}$ = $\text{-5.42m/s}$
(Remember that a square root is + or -, and the downward motion is negative.)

Now determine the time it takes for the book to land using the equation ${\text{v}}_{f}$ = ${\text{v}}_{i}$ + $g$$t$ and solving for $t$.
Since v"_i is 0m/s, the equation becomes v"_f = $g$$t$.

$t$ = ${\text{v}}_{f} / g$ = ${\text{-5.42m/s"/"-9.81m/s}}^{2}$ = 0.552s"

So, if you drop a book from a height of $\text{1.50m}$, such that only gravity is acting on it, the time it takes to land is $\text{0.552s}$, and its final velocity is $\text{-5.42m/s}$.

Example 2. Throwing a book down from a height of $\text{1.50m}$.
Lets say you threw a book down with an initial velocity of $\text{-2.00m/s}$ and an acceleration of ${\text{-10.50m/s}}^{2}$.
How long will it take the book to land, and what will its final velocity be?

Known/Given:
$a$ = ${\text{-10.50m/s}}^{2}$
$d$ = $\text{-1.50m}$
v"_i = $\text{-2.00m/s}$

Unknown:
time, $t$
final velocity v"_f

Equations:
${\text{v"_f}}^{2}$ = ${\text{v"_i}}^{2}$ + $\text{2} a d$
${\text{v}}_{f}$ = ${\text{v}}_{i}$ + $a t$

Solution:

First, determine the final velocity of the book when it lands using the kinematic equation ${\text{v"_f}}^{2}$ = ${\text{v"_i}}^{2}$ + $\text{2} a d$, and solving for v"_f.

${\text{v"_f}}^{2}$ = (${\text{-2.00m/s)}}^{2}$ + "("2")$\left({\text{-10.50m/s}}^{2}\right)$$\left(\text{-1.50m}\right)$ = ${\text{35.50m"^2"/s}}^{2}$

$\text{v"_f}$ = sqrt${\text{35.50m"^2"/s}}^{2}$ = $\text{-5.96m/s}$
(Again, remember that a square root is + or -, and here the downward motion is negative.)

Now determine the time it takes the thrown book to land using the equation ${\text{v}}_{f}$ = ${\text{v}}_{i}$ + $a t$, and solving for $t$.

$t$ = $\frac{{\text{v"_f - "v}}_{i}}{a}$ = $\text{-5.96m/s - (-2.00m/s)"/"-10.50m/s"^2}$ = $\text{0.337s}$

So a book thrown down from a height of $\text{1.50m}$ with an initial velocity of $\text{-2.00m/s}$ and an acceleration of $\text{-10.50m/s"^2}$, will land after 0.337s" with a final velocity of $\text{-5.96m/s}$. This is a shorter time and a faster final velocity than when the book was dropped.

Nov 15, 2014

Yes. When an object is dropped, its initial velocity is zero and its acceleration is that of gravity, which is ${\text{-9.81m/s}}^{2}$. When an object is thrown to the ground, its initial velocity is greater than zero and its acceleration is greater than ${\text{-9.81m/s}}^{2}$.

There are four kinematic equations that are used to study motion in one direction. .

Two of these equations will be used to test how fast an object falls, and its final velocity.

Note: When an object is dropped so that only the force of gravity is acting on it, the symbol for acceleration becomes $g$, and is ${\text{-9.81m/s}}^{2}$

Example 1. Dropping a book from a height of $\text{1.50}$ meters.
How long will it take the book to land, and what will its final velocity be?

Known/Given:
$g$ = ${\text{-9.81m/s}}^{2}$
$d$ = $\text{-1.50m}$
v"_i = $\text{0m/s}$

Unknown:
time, $t$
final velocity v"_f

Equations:
${\text{v"_f}}^{2}$ = ${\text{v"_i}}^{2}$ + $\text{2} g d$
${\text{v}}_{f}$ = ${\text{v}}_{i}$ + $g$$t$

Solution:
First, determine the final velocity of the book when it lands using the kinematic equation ${\text{v"_f}}^{2}$ = ${\text{v"_i}}^{2}$ + $\text{2} g d$, and solving for v"_f. Since the initial velocity is zero, the equation becomes

${\text{v"_f}}^{2}$ = $\text{2} g d$

v"_f = sqrt$\text{2} g d$ = sqrt$\left(\text{2}\right)$$\left({\text{-9.81m/s}}^{2}\right)$$\left(\text{-1.50m}\right)$ = sqrt${\text{29.43m"^2"/s}}^{2}$ = $\text{-5.42m/s}$
(Remember that a square root is + or -, and the downward motion is negative.)

Now determine the time it takes for the book to land using the equation ${\text{v}}_{f}$ = ${\text{v}}_{i}$ + $g$$t$ and solving for $t$.
Since v"_i is 0m/s, the equation becomes v"_f = $g$$t$.

$t$ = ${\text{v}}_{f} / g$ = ${\text{-5.42m/s"/"-9.81m/s}}^{2}$ = 0.552s"

So, if you drop a book from a height of $\text{1.50m}$, such that only gravity is acting on it, the time it takes to land is $\text{0.552s}$, and its final velocity is $\text{-5.42m/s}$.

Example 2. Throwing a book down from a height of $\text{1.50m}$.
Lets say you threw a book down with an initial velocity of $\text{-2.00m/s}$ and an acceleration of ${\text{-10.50m/s}}^{2}$.
How long will it take the book to land, and what will its final velocity be?

Known/Given:
$a$ = ${\text{-10.50m/s}}^{2}$
$d$ = $\text{-1.50m}$
v"_i = $\text{-2.00m/s}$

Unknown:
time, $t$
final velocity v"_f

Equations:
${\text{v"_f}}^{2}$ = ${\text{v"_i}}^{2}$ + $\text{2} a d$
${\text{v}}_{f}$ = ${\text{v}}_{i}$ + $a t$

Solution:

First, determine the final velocity of the book when it lands using the kinematic equation ${\text{v"_f}}^{2}$ = ${\text{v"_i}}^{2}$ + $\text{2} a d$, and solving for v"_f.

${\text{v"_f}}^{2}$ = (${\text{-2.00m/s)}}^{2}$ + "("2")$\left({\text{-10.50m/s}}^{2}\right)$$\left(\text{-1.50m}\right)$ = ${\text{35.50m"^2"/s}}^{2}$

$\text{v"_f}$ = sqrt${\text{35.50m"^2"/s}}^{2}$ = $\text{-5.96m/s}$
(Again, remember that a square root is + or -, and here the downward motion is negative.)

Now determine the time it takes the thrown book to land using the equation ${\text{v}}_{f}$ = ${\text{v}}_{i}$ + $a t$, and solving for $t$.

$t$ = $\frac{{\text{v"_f - "v}}_{i}}{a}$ = $\text{-5.96m/s - (-2.00m/s)"/"-10.50m/s"^2}$ = $\text{0.337s}$

So a book thrown down from a height of $\text{1.50m}$ with an initial velocity of $\text{-2.00m/s}$ and an acceleration of $\text{-10.50m/s"^2}$, will land after 0.337s"# with a final velocity of $\text{-5.96m/s}$. This is a shorter time and a faster final velocity than when the book was dropped.