Chapter 2, Problem 52P

A package is dropped from a helicopter that is descending steadily at a speed v_n. After t seconds have elapsed, (a) what is the speed of the package in terms of v_n, g, and t? (b) What distance d is it from the helicopter in terms of g and t? (c) What are the answers to parts (a) and (b) if the helicopter is rising steadily at the same speed?

To determine

The speed of the package.

Answer to Problem 52P

The speed of the package is given by the formula $\left|v_p\right|=v_0+gt$.

Explanation of Solution

Given Info: The initial velocity of the package is $-v_0$, the acceleration of the package is $-g$, and the time taken is $t$.

Explanation:

The time starts when the package leaves the helicopter. The initial velocity of the helicopter and the package is same.

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

$v_p=v_i+at$

Here,

$v_p$ is the final velocity of the package

$v_i$ is the initial velocity of the package

$a$ is the acceleration of the package

$t$ is the time taken

Substitute $-v_0$ for $v_i$, $t$ for $t$, and $-g$ for $a$ to find $v_p$.

$v_p=-v_0-gt$

Speed of the package is,

$\begin{array}{c}\left|v_p\right|=\left|-v_0-gt\right|\\ =v_0+gt\end{array}$

Thus, the speed of the package is given by the formula $\left|v_p\right|=v_0+gt$.

Conclusion:

The speed of the package is given by the formula $\left|v_p\right|=v_0+gt$.

To determine

The distance separating the package and the helicopter.

Answer to Problem 52P

The distance separating the package and the helicopter is given by the formula $d=\frac{1}{2}g{t}^2$.

Explanation of Solution

Given Info: The initial velocity of the helicopter is $-v_0$, the acceleration of the helicopter is $0$, the time taken is $t$.

Explanation:

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

${\left(\Delta y\right)}_p=v_{i}t+\frac{1}{2}a{t}^2$

Here,

${\left(\Delta y\right)}_p$ is the displacement of the package

Substitute $-v_0$ for $v_i$, $t$ for $t$ and $-g$ for $a$ to find ${\left(\Delta y\right)}_p$.

$\begin{array}{c}{\left(\Delta y\right)}_p=-v_{0}t+\frac{1}{2}\left(-g\right)t^2\\ =-\left(v_{0}t+\frac{1}{2}g{t}^2\right)\end{array}$

Thus, the displacement of the package is given by the equation

${\left(\Delta y\right)}_p=-\left(v_{0}t+\frac{1}{2}g{t}^2\right)$

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

${\left(\Delta y\right)}_h=v_{i}t+\frac{1}{2}{a}_h{t}^2$

Here,

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

$a_h$ is the acceleration of the helicopter

Substitute $-v_0$ for $v_i$, $t$ for $t$, and $0$ for $a$ to find ${\left(\Delta y\right)}_h$.

$\begin{array}{c}{\left(\Delta y\right)}_h=\left(-v_0\right)t+\frac{1}{2}\left(0\right)t^2\\ =-v_{0}t\end{array}$

Thus, the displacement of the helicopter is given by the equation ${\left(\Delta y\right)}_h=-v_{0}t$.

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

$d=\left|\left|{\left(\Delta y\right)}_p\right|-\left|{\left(\Delta y\right)}_h\right|\right|$

Here,

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

Substitute $-\left(v_{0}t+\frac{1}{2}g{t}^2\right)$ for ${\left(\Delta y\right)}_p$ and $-v_{0}t$ for ${\left(\Delta y\right)}_h$ to find $d$.

$\begin{array}{c}d=\left|\left|-\left(v_{0}t+\frac{1}{2}g{t}^2\right)\right|-\left|-v_{0}t\right|\right|\\ =\frac{1}{2}g{t}^2\end{array}$

Thus, the distance separating the package and the helicopter is given by the formula $d=\frac{1}{2}g{t}^2$.

Conclusion:

The distance separating the package and the helicopter is given by the formula $d=\frac{1}{2}g{t}^2$.

To determine

Answers to parts (a) and (b) if the helicopter is rising steadily at the same speed.

Answer to Problem 52P

Answers to parts (a) and (b) if the helicopter is rising steadily at the same speed are $\left|v_p\right|=v_0+gt$ and $d=\left(1/2\right)g{t}^2$.

Explanation of Solution

Given Info: The initial velocity of the package is $v_0$, the acceleration of the package is $-g$, the time taken is $t$, and the acceleration of the helicopter is $0$.

Explanation:

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

$v_p=v_i+at$

Substitute $v_0$ for $v_i$, $t$ for $t$, and $-g$ for $a$ to find $v_p$.

$v_p=v_0-gt$

Speed of the package is,

$\left|v_p\right|=\left|v_0-gt\right|$

Thus, the speed of the package is given by the formula $\left|v_p\right|=\left|v_0-gt\right|$.

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

${\left(\Delta y\right)}_p=v_{i}t+\frac{1}{2}a{t}^2$

Here

${\left(\Delta y\right)}_p$ is the displacement of the package

Substitute $v_0$ for $v_i$, $t$ for $t$, and $-g$ for $a$ to find ${\left(\Delta y\right)}_p$.

$\begin{array}{c}{\left(\Delta y\right)}_p=v_{0}t+\frac{1}{2}\left(-g\right)t^2\\ =v_{0}t-\frac{1}{2}g{t}^2\end{array}$

Thus, the displacement of the package is given by the equation ${\left(\Delta y\right)}_p=v_{0}t-\frac{1}{2}g{t}^2$.

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

${\left(\Delta y\right)}_h=v_{i}t+\frac{1}{2}{a}_h{t}^2$

Here,

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

$a_h$ is the acceleration of the helicopter

Substitute $v_0$ for $v_i$, $t$ for $t$, and $0$ for $a$ to find ${\left(\Delta y\right)}_h$.

$\begin{array}{c}{\left(\Delta y\right)}_h=v_{0}t+\frac{1}{2}\left(0\right)t^2\\ =v_{0}t\end{array}$

Thus, the displacement of the helicopter is given by the equation ${\left(\Delta y\right)}_h=v_{0}t$.

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

$d=\left|\left|{\left(\Delta y\right)}_p\right|-\left|{\left(\Delta y\right)}_h\right|\right|$

Here,

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

Substitute $v_{0}t-\frac{1}{2}g{t}^2$ for ${\left(\Delta y\right)}_p$ and $v_{0}t$ for ${\left(\Delta y\right)}_h$ to find $d$.

$\begin{array}{c}d=\left|\left|\left(v_{0}t-\frac{1}{2}g{t}^2\right)\right|-\left|v_{0}t\right|\right|\\ =\frac{1}{2}g{t}^2\end{array}$

Thus, the distance separating the package and the helicopter is given by the formula $d=\frac{1}{2}g{t}^2$.

Conclusion:

Answers to parts (a) and (b) if the helicopter is rising steadily at the same speed are $\left|v_p\right|=v_0+gt$ and $d=\left(1/2\right)g{t}^2$.