Chapter 2, Problem 78QAP

To determine

(a)

The speed of payload at the end of the stage 1.

Answer to Problem 78QAP

The speed of payload at the end of the stage 1 is 150 m/s

Explanation of Solution

Given:

Acceleration in stage 1, $a_1=15.0{\text{ m/s}}^{\text{2}}$

Acceleration in stage 2, $a_2=12.0{\text{ m/s}}^{\text{2}}$

Time to burn payload in stage 1, $t_1=10\text{ sec}$

Time to burn payload in stage 2, $t_2=5\text{ sec}$

Formula used:

  $v=u+at$

u = initial velocity

t = time

a = acceleration

Calculation:

To find the velocity at the end of first stage, substitute the values in above equation,

  $\begin{array}{c}{v}_1=u+a_1{t}_1\\ \text{= 0+(15}{\text{.0 m/s}}^{\text{2}}\times 1\text{0 sec)}\\ \text{= 150 m/s}\end{array}$

To determine

(b)

The speed of payload at the end of the stage 2.

Answer to Problem 78QAP

The speed of payload at the end of the stage 2 is 210 m/s

Explanation of Solution

Given:

Acceleration in stage 1, $a_1=15.0{\text{ m/s}}^{\text{2}}$

Acceleration in stage 2, $a_2=12.0{\text{ m/s}}^{\text{2}}$

Time to burn payload in stage 1, $t_1=10\text{ sec}$

Time to burn payload in stage 2, $t_2=5\text{ sec}$

Formula used:

  $v=u+at$

u = initial velocity

t = time

a = acceleration

Calculation:

To find the velocity at the end of second stage, substitute the values in above equation,

  $\begin{array}{c}{v}_2=v_1+a_2{t}_2\\ \text{= 150+(12}{\text{.0 m/s}}^{\text{2}}\times \text{5 seconds)}\\ \text{= 210 m/s}\end{array}$

To determine

(c)

The altitude of the rocket at the end of stage 1.

Answer to Problem 78QAP

The altitude of the rocket at the end of stage 1 is 750 m.

Explanation of Solution

Given:

Acceleration in stage 1, $a_1=15.0{\text{ m/s}}^{\text{2}}$

Acceleration in stage 2, $a_2=12.0{\text{ m/s}}^{\text{2}}$

Time to burn payload in stage 1, $t_1=10\text{ sec}$

Time to burn payload in stage 2, $t_2=5\text{ sec}$

Formula used:

  $s=ut+\frac{1}{2}a{t}^2$

s = distance travelled.

u = initial velocity

t = time

a = acceleration

Calculation:

Substituting the values in above equation, we get-

  $\begin{array}{c}{s}_1=u{t}_1+\frac{1}{2}{a}_1{t}_1{}^2\\ \text{= 0 + }\frac{\text{1}}{\text{2}}\times \text{(15}{\text{.0 m/s}}^{\text{2}}\text{)}\times {\text{(10 sec)}}^{\text{2}}\\ \text{= 750m}\end{array}$

To determine

(d)

The altitude of the rocket at the end of stage 2.

Answer to Problem 78QAP

The altitude of the rocket at the end of stage 2 is 1650 m

Explanation of Solution

Given:

Acceleration in stage 1, $a_1=15.0{\text{ m/s}}^{\text{2}}$

Acceleration in stage 2, $a_2=12.0{\text{ m/s}}^{\text{2}}$

Time to burn payload in stage 1, $t_1=10\text{ sec}$

Time to burn payload in stage 2, $t_2=5\text{ sec}$

Formula used:

  $s=ut+\frac{1}{2}a{t}^2$

s = distance travelled.

u = initial velocity

t = time

a = acceleration

Calculation:

Substituting the values in above equation, we get-

  $\begin{array}{c}{s}_2=s_1+v_1{t}_2+\frac{1}{2}{a}_2{t}_2{}^2\\ \text{= 750+}\left(\text{150}\times \text{5}\right)\text{+ }\frac{\text{1}}{\text{2}}\times \text{(12}\text{.0)}\times {\text{5}}^{\text{2}}\\ \text{= 1650 m }\end{array}$

To determine

(e)

The maximum altitude obtained.

Answer to Problem 78QAP

The maximum altitude obtained is 3900 m.

Explanation of Solution

Given:

Acceleration in stage 1, $a_1=15.0{\text{ m/s}}^{\text{2}}$

Acceleration in stage 2, $a_2=12.0{\text{ m/s}}^{\text{2}}$

Time to burn payload in stage 1, $t_1=10\text{ sec}$

Time to burn payload in stage 2, $t_2=5\text{ sec}$

Formula used:

  $v^2=u^2-2g{h}_3$

v = final velocity.

u = initial velocity.

g = acceleration due to gravity.

  $h_3$ = the height after stage 2

Calculation:

The extra height covered after stage 2 is,

  $\begin{array}{c}{v}_3{}^2=v_2{}^2-2g{s}_3\\ 0={210}^2-2\times 9.81\times s_3\\ s_3=\frac{{210}^2}{2\times 9.81}\\ =2247.706\text{m}\end{array}$

So, the highest distance is,

  $\begin{array}{c}H=s_2+s_3\\ =1650+2247.706\\ =3897.7\\ \approx 3900\text{m}\end{array}$

To determine

(f)

The total travel time of the payload from launch to landing.

Answer to Problem 78QAP

The total travel time of the payload from launch to landing is 64.64 sec.

Explanation of Solution

Given:

Acceleration in stage 1, $a_1=15.0{\text{ m/s}}^{\text{2}}$

Acceleration in stage 2, $a_2=12.0{\text{ m/s}}^{\text{2}}$

Time to burn payload in stage 1, $t_1=10\text{ sec}$

Time to burn payload in stage 2, $t_2=5\text{ sec}$

Formula used:

  $s=ut+\frac{1}{2}a{t}^2$

s = distance travelled.

u = initial velocity

t = time

a = acceleration

Calculation:

Time taken to cover extra height is,

  $\begin{array}{c}{v}_3=v_2-g{t}_3\\ 0=210-9.81t_3\\ t_3=\frac{210}{9.81}\\ \approx 21.43\text{sec}\end{array}$

Time taken to drop back to ground is,

  $\begin{array}{c}H=u{t}_4+\frac{1}{2}g{t}_4{}^2\\ t_4=\sqrt{\frac{2H}{g}}\\ =\sqrt{\frac{2\times 3900}{9.8}}\\ =28.21\text{sec}\end{array}$

So, total time of flight is,

  $\begin{array}{c}t=t_1+t_2+t_3+t_4\\ =10+5+21.43+28.21\\ =64.64\text{sec}\end{array}$