Chapter 2, Problem 68P

(a)

To determine

What is the rocket’s altitude when the engine fails?

Answer to Problem 68P

Rocket’s altitude when the engine fails is $25\text{km}$.

Explanation of Solution

Write the equation to find the displacement of rocket.

$s=ut+\frac{1}{2}a{t}^2$                   (I)

Here, $u$ is the initial velocity, $s$ is the altitude, $t$ is the time, $a$ is the acceleration of the rocket.

The initial velocity of rocket is zero, therefore, rewrite equation (I)

$s=\frac{1}{2}a{t}^2$ (II)

Conclusion

Substitute $20\text{m}/{\text{s}}^{\text{2}}$ for $a$ and $50\text{s}$ for $t$ in equation (II) to get $s$.

$\begin{array}{c}s=\frac{1}{2}\left(20\text{m}/{\text{s}}^{\text{2}}\right){\left(50\text{s}\right)}^2\\ =25\text{km}\end{array}$

Therefore, Rocket’s altitude when the engine fails is $25\text{km}$.

(b)

To determine

When does the rocket reach the maximum height?

Answer to Problem 68P

The rocket will reach maximum height at $152\text{s}$.

Explanation of Solution

Write the equation to find the final velocity of the rocket.

$\begin{array}{c}{v}_y=v_{iy}-g\Delta t\\ =a\Delta t_1-g\Delta t\end{array}$ (I)

Here, $v_y$ is the final velocity of rocket, $a$ is the acceleration, $\Delta t_1$ is the change in time, $g$ is the acceleration due to gravity

The final velocity of rocket is zero. Hence rewrite equation (I).

$\begin{array}{l}0=a\Delta t_1-g\Delta t\\ \Delta t=\frac{a}{g}\Delta t_1\end{array}$ (II)

Conclusion:

Substitute $20\text{m}/{\text{s}}^{\text{2}}$ for $a$ , $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ , and $50\text{s}$ for $\Delta t_1$ in  equation (II) to get $\Delta t$.

$\begin{array}{c}\Delta t=\frac{\left(20\text{m}/{\text{s}}^{\text{2}}\right)}{\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}\left(50\text{s}\right)\\ =102\text{s}\end{array}$

Therefore, The rocket will reach maximum height at $152\text{s}$.

(c)

To determine

What is the maximum height reached by the rocket?

Answer to Problem 68P

The maximum height reached by rocket is $76\text{km}$.

Explanation of Solution

Write the equation to find the maximum height reached by the rocket.

$y_f=y_i+v_{iy}\Delta t-\frac{1}{2}g{\left(\Delta t\right)}^2$ (I)

Here, $y_f$ is the maximum height, $y_i$ is the initial height, $v_{iy}$ is the initial velocity, $\Delta t$ is the change in time, $g$ is acceleration due to gravity

Substitute $\frac{a}{g}\Delta t_1$ for $\Delta t$ , and $a\Delta t$ for $v_{iy}$ in equation (I).

$y_f=y_i+\left(a\Delta t_1\right)\left(\frac{a}{g}\Delta t_1\right)-\frac{1}{2}g{\left(\frac{a}{g}\Delta t_1\right)}^2$ (II)

Solve equation (II) to get $y_f$.

$\begin{array}{c}{y}_f=y_i+\frac{a^2{\left(\Delta t_1\right)}^2}{g}-\frac{a^2{\left(\Delta t_1\right)}^2}{2g}\\ =y_i+\frac{a^2{\left(\Delta t_1\right)}^2}{2g}\end{array}$ (III)

Conclusion:

Substitute $25\text{km}$ for $y_i$ , $20\text{m}/{\text{s}}^{\text{2}}$ for $a$ , $50\text{s}$ for $\Delta t$ , $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (III) to get $y_f$.

$\begin{array}{c}{y}_f=25\text{km+}\frac{{\left(20\text{m}/{\text{s}}^{\text{2}}\right)}^2{\left(50\text{s}\right)}^2}{2\left(9.8\text{N}/\text{kg}\right)}\\ =76\text{km}\end{array}$

Therefore, The maximum height reached by rocket is $76\text{km}$.

(d)

To determine

What is the velocity of the rocket just before it hits the ground?

Answer to Problem 68P

The velocity of the rocket just before it hits the ground is $1220\text{m}/\text{s}$ downwards.

Explanation of Solution

Write the equation to find the final velocity of rocket.

$v_{fy}{}^2=v_{iy}{}^2-2g\Delta y$ (I)

Here, $v_{fy}$ is the final velocity, $v_{iy}$ is the initial velocity, $g$ is the acceleration due to gravity, $\Delta y$ is the displacement

The initial velocity of the rocket is zero. Hence rewrite equation (I) to get $v_{fy}$.

$v_{fy}=\sqrt{-2g\Delta y}$ (II)

Conclusion:

Substitute $9.\text{8}\text{m}/{\text{s}}^{\text{2}}$ for $g$ , and $76\text{km}$ for $\Delta y$ in equation (II) to get $v_{fy}$.

$\begin{array}{c}{v}_{fy}=\sqrt{-2\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(0-76\times {10}^3\text{m}\right)}\\ =1220\text{m}/\text{s}\end{array}$

Therefore, The velocity of the rocket just before it hits the ground is $1220\text{m}/\text{s}$ downwards.