Chapter 2, Problem 82QAP

To determine

(a)

The maximum altitude of the rocket.

Answer to Problem 82QAP

The maximum altitude of the rocket is $1091.7\text{m}$

Explanation of Solution

Given:

a₁ = $\text{2}{\text{.00 m/s}}^{\text{2}}$

a₂ = $\text{3}{\text{.00 m/s}}^{\text{2}}$

t₁ = 15 sec

t₂ = 12 sec

Formula used:

  $H=ut+\frac{1}{2}a{t}^2$

S = distance travelled

u = initial velocity

a = acceleration

t = time taken

Calculation:

During first stage, the height reached is,

  $\begin{array}{c}{H}_1=u{t}_1+\frac{1}{2}{a}_1{t}_1{}^2\\ =0\times 4+\frac{1}{2}\times 2\times {16}^2\\ =256\text{m}\end{array}$

At the end of first stage speed is,

  $\begin{array}{c}{v}_1=u+a_1{t}_1\\ =0+2\times 16\\ =32\text{m}/\text{s}\end{array}$

During second stage, the height reached is,

  $\begin{array}{c}{H}_2=v_1{t}_2+\frac{1}{2}{a}_2{t}_2{}^2\\ =32\times 12+\frac{1}{2}\times 3\times {12}^2\\ =600\text{m}\end{array}$

At the end of second stage speed is,

  $\begin{array}{c}{v}_2=v_1+a_2{t}_2\\ =32+3\times 12\\ =68\text{m}/\text{s}\end{array}$

At the end of second stage, the rocket will be under the effect of gravity, so it will continue to move upwards up to a point where its kinetic energy will be zero.

The time duration for reaching the maximum altitude from the end of second stage is,

  $\begin{array}{l}{v}_3=v_2-g{t}_3\\ 0=68-\left(9.8\times t_3\right)\\ t_3=\frac{68}{9.8}=6.938\text{sec}\end{array}$

The distance between the end of second stage and maximum altitude is,

  $\begin{array}{c}{H}_3=v_2{t}_3-\frac{1}{2}g{t}_3{}^2\\ =68\times 6.938-\frac{1}{2}\times 9.81\times {\left(6.938\right)}^2\\ =235.7\text{m}\end{array}$

So, the maximum height is,

  $\begin{array}{c}{H}_1=H_1+H_2+H_3\\ =256+600+235.7\\ =1091.7\text{m}\end{array}$

To determine

(b)

The average speed and average velocity when the rocket fall back to launch pad.

Answer to Problem 82QAP

The average speed is $44.69\text{m}/\text{s}$ and average velocity is 0.

Explanation of Solution

Given:

The maximum altitude is $1091.67\text{m}$

Calculation:

Once rocket reaches the maximum height, it will start to fall and reach the launch pad on Earth surface.

The time duration taken by rocket to fall back is,

  $\begin{array}{l}H=v_3{t}_4+\frac{1}{2}\times g\times t_4{}^2\\ 1091.7\text{m}=0\times t_4+\frac{1}{2}\times 9.81\times t_4{}^2\\ t_4=14.91\sec \end{array}$

Total time duration of flight,

  $\begin{array}{c}t=t_1+t_2+t_3+t_4\\ =15+12+6.938+14.91\\ =48.85\text{sec}\end{array}$

Total distance travelled is,

  $\begin{array}{c}{H}_t=H+H\\ =2H\\ =2\times 1091.7\text{m}\\ \text{=2183}\text{.4}\text{m}\end{array}$

So average speed will be,

  $\begin{array}{c}S=\frac{2183.34}{48.85}\\ =44.69\text{m}/\text{s}\end{array}$

Since, the initial and final position of rocket is same, so the net displacement is zero. Therefore, the average velocity will be,

  $\begin{array}{c}v=\frac{Di}{t}\\ =\frac{0}{t}\\ =0\text{m}/\text{s}\end{array}$