Chapter 14.3, Problem 14.100P

A rocket weighs 2600 lb. including 2200 lb of fuel, which is consumed at the rate of 25 lb/s and ejected with a relative velocity of 13.000 ft/s. Knowing that the rocket is fired vertically from the ground, determine (a) its acceleration as it is fired, (b) its acceleration as the last particle of fuel is being consumed, (c) the altitude at which all the fuel has been consumed. (d) the velocity of the rocket at that time.

To determine

The acceleration as it is fired of the rocket.

Answer to Problem 14.100P

The value of acceleration is a=92.8ft/s²

Explanation of Solution

The weight of the rocket as:

$W_R=mg$

Substitute $\begin{array}{l}{W}_R=2600Ib\\ g=32.2ft/s^2\end{array}$ we get

$\begin{array}{l}2600Ib=m\left(32.2ft/s^2\right)\\ m=\frac{2600}{32.2}\\ =80.75Ib.s^2/ft\end{array}$

Substitute required values, we get:

$a=\frac{10.088\times {10}^3}{80.75}-32.2ft/s^2$

$\begin{array}{l}=124.93-32.2\\ =92.8ft/s^2\end{array}$

To determine

The acceleration as the last particle of fuel is consumed.

Answer to Problem 14.100P

The value of acceleration as the last particle of fuel is a=780ft/s².

Explanation of Solution

Find the weight of the rocket the condition is fuel is consumed.

$W_o=W_R-W_F$

Substituting all required values, we get:

$\begin{array}{l}{W}_o=2600Ib-2200Ib\\ =400Ib\end{array}$

The weight of the rocket is:

$\begin{array}{l}{W}_o=mg\\ 400Ib=m\left(32.2\right)\\ m=\frac{400}{32.2}\\ m=12.422Ib.s^2/ft\end{array}$

Substitute all the values, we get:

$\begin{array}{l}a=\frac{P}{m}-g\\ a=\frac{10.088\times {10}^{3}Ib}{12.422Ib.s^2/ft}-32.2ft/s^2\\ =812.2-32.2\\ =779.9ft/s^2\\ =780ft/s^2\end{array}$

To determine

The altitude at which all the fuel has been consumed.

Answer to Problem 14.100P

The value of altitude at which all the fuel has been consumed is $h=119.3mi.$

Explanation of Solution

Formulate the relation for final velocity of the rocket is:

$\begin{array}{l}v=u\ln \left(\frac{m_o}{m_o-\left(qt\right)}\right)-gt\\ \frac{dy}{dt}=u\ln \left(\frac{m_o}{m_o-\left(qt\right)}\right)-gt---(1)\\ dy=\left[u\ln \left(\frac{m_o}{m_o-\left(qt\right)}\right)-gt\right]dt\end{array}$

Consider the relation for the altitude of the rocket:

$h=\int dy$

Here $dy=\left[u\ln \left(\frac{m_o}{m_o-\left(qt\right)}\right)-gt\right]dt$

$\begin{array}{l}h=\underset{0}{\overset{t}{\int }}\left(\left[u\ln \left(\frac{m_o}{m_o-\left(qt\right)}\right)-gt\right]dt\right)\\ =-u\underset{0}{\overset{t}{\int }}\ln \frac{m_o-qt}{m_o}dt-\frac{1}{2}g{t}^2---(2)\end{array}$

Here $z=\frac{m_o-qt}{m_o}$

So,

$\begin{array}{l}dz=-\frac{q}{m_o}dt\\ dt=-\frac{m_o}{q}dz\end{array}$

Substitute these values, we get:

$\begin{array}{l}h=-u\underset{0}{\overset{t}{\int }}\ln z\left(-\frac{m_o}{q}dz\right)-\frac{1}{2}g{t}^2\\ =\frac{m_{o}u}{q}\int \ln zdz-\frac{1}{2}g{t}^2\\ =\frac{m_{o}u}{q}{\left[\left(z\ln z-z\right)\right]}_{z_o}^z-\frac{1}{2}g{t}^2\end{array}$

Substitute $z=\frac{m_o-qt}{m_o}$ in equation 2, we get:

$\begin{array}{l}h=\frac{m_{o}u}{q}\left[\frac{m_o-qt}{m_o}\left(\ln \frac{m_o-qt}{m_o}-1\right)-\frac{m_o}{m_o}\left(\ln \frac{m_o}{m_o}-1\right)\right]-\frac{1}{2}g{t}^2\\ =\frac{m_{o}u}{q}\left[\left(1-\frac{q}{m_o}t\right)\times \ln \frac{m_o-qt}{m_o}-\left(1-\frac{q}{m_o}t\right)-\frac{m_o}{m_o}\left(\ln \frac{m_o}{m_o}-1\right)\right]-\frac{1}{2}g{t}^2\end{array}$

$\begin{array}{l}=\frac{m_{o}u}{q}\left[\ln \frac{m_o-qt}{m_o}-\left(\frac{q}{m_o}t\times \ln \frac{m_o-qt}{m_o}\right)+\frac{q}{m_o}t\right]-\frac{1}{2}g{t}^2\\ =\left(\frac{m_{o}u}{q}\times \ln \frac{m_o-qt}{m_o}\right)-\left(\frac{m_{o}u}{q}\times \frac{q}{m_o}t\times \ln \frac{m_o-qt}{m_o}\right)+\left(\frac{m_{o}u}{q}\times \frac{q}{m_o}t\right)-\frac{1}{2}g{t}^2\end{array}$

$=\left(\frac{m_{o}u}{q}\times \ln \frac{m_o-qt}{m_o}\right)-\left(ut\times \ln \frac{m_o-qt}{m_o}\right)+\left(ut\right)-\frac{1}{2}g{t}^2----(3)$

Find the time t by using below formula, we get:

$\begin{array}{l}t=\frac{W_f}{q}\\ t=\frac{2200}{25}\\ =88s\end{array}$

Substitute all the values, we get:

$\begin{array}{l}h=ut+\left(\frac{m_{o}u}{q}-ut\right)\ln \frac{m_o-qt}{m_o}-\frac{1}{2}g{t}^2\\ =\left(13000\times 88\right)+\left[\left(\frac{2600\times 13000}{25}-\left(13000\times 88\right)\right)\ln \frac{2600-2200}{2600}\right]-\frac{1}{2}\left(32.2\right){\left(88\right)}^2\\ =1144000-389334.85-124678.4\\ =629986.75ft\times \frac{1mile}{5280ft}\\ h=119.3mi\end{array}$

To determine

The velocity of the rocket.

Answer to Problem 14.100P

The value of velocity at that time is V$=14660mi/h.$

Explanation of Solution

Substitute all the required values in equation 1, we get:

$\begin{array}{l}v=13000\ln \left(\frac{2600}{2600-2200}\right)-\left(32.2\right)\left(88\right)\\ =21500ft/s\times \frac{1mile}{5280ft}\times \frac{3600s}{1hr}\\ =14660mi/h\end{array}$