Chapter 2, Problem 50P

A small mailbag is released from a helicopter that is descending steadily at 1.50 m/s. After 2.00 s, (a) what is the speed of the mailbag, and (b) how far is it below the helicopter? (c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 1.50 m/s?

To determine

The magnitude of the velocity of the mailbag.

Answer to Problem 50P

The magnitude of the velocity of the mailbag is $21.1\text{m/s}$.

Explanation of Solution

Given Info: The initial velocity of the mailbag is $-1.50\text{m/s}$, the acceleration of the mailbag is $-9.80{\text{m/s}}^2$, and the time taken is $2.00\text{s}$.

Explanation:

The formula used to calculate final velocity of the mailbag is,

$v_{bag}=v_0+at$

Here,

$v_{bag}$ is the final velocity of the mailbag

$v_0$ is the initial velocity of the mailbag

$a$ is the acceleration of the mailbag

$t$ is the time taken

Substitute $-1.50\text{m/s}$ for $v_0$, $2.00\text{s}$ for $t$, and $-9.80{\text{m/s}}^{\text{2}}$ for $a$ to find $v_{bag}$.

$\begin{array}{c}{v}_{bag}=-1.50\text{m/s}+\left(-9.80{\text{m/s}}^{\text{2}}\right)\left(2.00\text{s}\right)\\ =-21.1\text{m/s}\end{array}$

Thus, the magnitude of the velocity of the mailbag is $21.1\text{m/s}$.

Conclusion:

The magnitude of the velocity of the mailbag is $21.1\text{m/s}$.

To determine

The distance separating the package and the helicopter.

Answer to Problem 50P

The distance separating the package and the helicopter is $19.6\text{m}$.

Explanation of Solution

The formula used to calculate the displacement of the mailbag is,

${\left(\Delta y\right)}_{bag}=\left(\frac{v_{bag}+v_0}{2}\right)t$

Here,

${\left(\Delta y\right)}_{bag}$ is the displacement of the mailbag

Substitute $-1.50\text{m/s}$ for $v_0$, $2.00\text{s}$ for $t$, and $-21.1\text{m/s}$ for $v_{bag}$ to find ${\left(\Delta y\right)}_{bag}$.

$\begin{array}{c}{\left(\Delta y\right)}_{bag}=\left(\frac{\left(-21.1\text{m/s}\right)+\left(-1.50\text{m/s}\right)}{2}\right)\left(2.00\text{s}\right)\\ =-22.6\text{m}\end{array}$

Thus, the displacement of the mailbag is $-22.6\text{m}$.

The formula used to calculate the displacement of the copter is,

${\left(\Delta y\right)}_{copter}=v_{0}t$

Here,

${\left(\Delta y\right)}_{copter}$ is the displacement of the helicopter

Substitute $-1.50\text{m/s}$ for $v_0$ and $2.00\text{s}$ for $t$ to find ${\left(\Delta y\right)}_{copter}$.

$\begin{array}{c}{\left(\Delta y\right)}_{copter}=\left(-1.50\text{m/s}\right)\left(2.00\text{s}\right)\\ =-3.00\text{m}\end{array}$

Thus, the displacement of the copter is $-3.00\text{m}$.

The formula used to calculate the distance separating the package and the helicopter is,

$d=\left|\left|{\left(\Delta y\right)}_{bag}\right|-\left|{\left(\Delta y\right)}_{copter}\right|\right|$

Here,

$d$ is the distance separating the package and the helicopter

Substitute $-22.6\text{m}$ for ${\left(\Delta y\right)}_{bag}$ and $-3.00\text{m}$ for ${\left(\Delta y\right)}_{copter}$ to find $d$.

$\begin{array}{c}d=\left|\left|-22.6\text{m}\right|-\left|-3.00\text{m}\right|\right|\\ =19.6\text{m}\end{array}$

Thus, the distance separating the package and the helicopter is $19.6\text{m}$.

Conclusion:

The distance separating the package and the helicopter is $19.6\text{m}$.

To determine

Answers to part (a) and (b) if the helicopter is rising steadily at $1.50\text{m/s}$.

Answer to Problem 50P

Answers to part (a) and (b) if the helicopter is rising steadily at $1.50\text{m/s}$ are $18.1\text{m/s}$ and $19.6\text{m}$.

Explanation of Solution

Given Info: The initial velocity of the mailbag is $1.50\text{m/s}$, the acceleration of the mailbag is $-9.80{\text{m/s}}^2$, and the time taken is $2.00\text{s}$.

Explanation:

The formula used to calculate final velocity of the mailbag is,

$v_{bag}=v_0+at$

Here,

$v_{bag}$ is the final velocity of the mailbag

$v_0$ is the initial velocity of the mailbag

$a$ is the acceleration of the mailbag

$t$ is the time taken

Substitute $1.50\text{m/s}$ for $v_0$, $2.00\text{s}$ for $t$, and $-9.80{\text{m/s}}^{\text{2}}$ for $a$ to find $v_{bag}$.

$\begin{array}{c}{v}_{bag}=1.50\text{m/s}+\left(-9.80{\text{m/s}}^{\text{2}}\right)\left(2.00\text{s}\right)\\ =-18.1\text{m/s}\end{array}$

Thus, the magnitude of velocity of the mailbag is $18.1\text{m/s}$.

The formula used to calculate the displacement of the mailbag is,

${\left(\Delta y\right)}_{bag}=\left(\frac{v_{bag}+v_0}{2}\right)t$

Here,

${\left(\Delta y\right)}_{bag}$ is the displacement of the mailbag

Substitute $1.50\text{m/s}$ for $v_0$, $2.00\text{s}$ for $t$, and $-18.1\text{m/s}$ for $v_{bag}$ to find ${\left(\Delta y\right)}_{bag}$.

$\begin{array}{c}{\left(\Delta y\right)}_{bag}=\left(\frac{\left(-18.1\text{m/s}\right)+\left(1.50\text{m/s}\right)}{2}\right)\left(2.00\text{s}\right)\\ =-16.6\text{m}\end{array}$

Thus, the displacement of the mailbag is $-16.6\text{m}$.

The formula used to calculate the displacement of the copter is,

${\left(\Delta y\right)}_{copter}=v_{0}t$

Here,

${\left(\Delta y\right)}_{copter}$ is the displacement of the helicopter

Substitute $1.50\text{m/s}$ for $v_0$ and $2.00\text{s}$ for $t$ to find ${\left(\Delta y\right)}_{copter}$.

$\begin{array}{c}{\left(\Delta y\right)}_{copter}=\left(1.50\text{m/s}\right)\left(2.00\text{s}\right)\\ =3.00\text{m}\end{array}$

Thus, The displacement of the copter is $3.00\text{m}$.

The formula used to calculate the distance separating the package and the helicopter is,

$d=\left|{\left(\Delta y\right)}_{bag}-{\left(\Delta y\right)}_{copter}\right|$

Here,

$d$ is the distance separating the package and the helicopter

Substitute $-16.6\text{m}$ for ${\left(\Delta y\right)}_{bag}$ and $3.00\text{m}$ for ${\left(\Delta y\right)}_{copter}$ to find $d$.

$\begin{array}{c}d=\left|-16.6\text{m}-3.00\text{m}\right|\\ =19.6\text{m}\end{array}$

Thus, the distance separating the package and the helicopter is $19.6\text{m}$.

Conclusion:

Answers to part (a) and (b) if the helicopter is rising steadily at $1.50\text{m/s}$ are $18.1\text{m/s}$ and $19.6\text{m}$.