Chapter 2, Problem 49P

(a)

To determine

The time interval for which the rocket is in motion above the ground.

Answer to Problem 49P

The time interval for which the rocket is in motion above the ground is $41.0\text{ s}$.

Explanation of Solution

Initially the rocket moves vertically upward with initial speed $80.0\text{ m/s}$ and acceleration $4.00{\text{ m/s}}^2$ . Then the engine fails and the rocket starts to move upward with acceleration equal to the negative of acceleration due to gravity until it reaches the maximum altitude. After this the rocket comes back to ground with acceleration equal to the negative of acceleration due to gravity Take the ground level to be point $0$, point $1$ to be at the end of the engine burn, point $2$ to be the highest altitude reached by the rocket and point $3$ to be at just before the impact on the ground.

The diagram is shown below.

Write the kinematics equation.

  $v_f^2=v_i^2+2a\Delta y$        (I)

Here, $v_f$ is the final speed, $v_i$ is the initial speed, $a$ is the acceleration and $\Delta y$ is the displacement.

Apply equation (I) for the motion of the rocket from point $0$ to point $1$.

  $v_{f_{0-1}}^2=v_{i_{0-1}}^2+2a_{0-1}\Delta y_{0-1}$

Here, $v_{f_{0-1}}$ is the speed rocket when the engine fail, $v_{i_{0-1}}$ is the initial speed of the rocket at the launch, $a_{0-1}$ is the acceleration produced by engine thrust and $\Delta y_{0-1}$ is the displacement from point $0$ to point $1$ .

Substitute $80.0\text{ m/s}$ for $v_{i_{0-1}}$, $4.00{\text{ m/s}}^2$ for $a_{0-1}$ and $1000\text{ m}$ for $\Delta y_{0-1}$ in the above equation to find $v_{f_{0-1}}$.

  $\begin{array}{c}{v}_{f_{0-1}}^2={\left(80.0\text{ m/s}\right)}^2+2\left(4.00{\text{ m/s}}^2\right)\left(1000\text{ m}\right)\\ =14400{\text{ m}}^2{\text{/s}}^2\\ v_{f_{0-1}}=120\text{ m/s}\end{array}$

Write the kinematics equation.

  $v_f=v_i+at$        (II)

Here, $t$ is the time interval.

Apply equation (II) for the motion of the rocket from point $0$ to point $1$ .

  $v_{f_{0-1}}=v_{i_{0-1}}+a_{0-1}{t}_{0-1}$

Here, $t_{0-1}$ is the time taken by the rocket to reach point $1$ from point $0$ .

Rewrite the above equation for $t_{0-1}$.

  $t_{0-1}=\frac{v_{f_{0-1}}-v_{i_{0-1}}}{a_{0-1}}$

Substitute $120\text{ m/s}$ for $v_{f_{0-1}}$, $80.0\text{ m/s}$ for $v_{i_{0-1}}$ and $4.00{\text{ m/s}}^2$ for $a_{0-1}$ in the above equation to find $t_{0-1}$.

  $\begin{array}{c}{t}_{0-1}=\frac{120\text{ m/s}-80.0\text{ m/s}}{4.00{\text{ m/s}}^2}\\ =10.0\text{ s}\end{array}$

Apply equation (I) for the motion of the rocket from point $1$ to point $2$.

  $\begin{array}{c}{v}_{f_{1-2}}^2=v_{f_{0-1}}^2+2\left(-g\right)\Delta y_{1-2}\\ =v_{f_{0-1}}^2-2g\Delta y_{1-2}\end{array}$

Here, $v_{f_{1-2}}$ is the speed rocket at the maximum altitude, $g$ is the acceleration due to gravity and $\Delta y_{1-2}$ is the displacement from point $1$ to point $2$.

Rewrite the above equation for $\Delta y_{1-2}$ .

  $\Delta y_{1-2}=\frac{v_{f_{1-2}}^2-v_{f_{0-1}}^2}{-2g}$

At the maximum altitude the speed of the rocket will be zero.

Substitute $0$ for $v_{f_{1-2}}$, $120\text{ m/s}$ for $v_{f_{0-1}}$ and $9.80{\text{ m/s}}^2$ for $g$ in the above equation to find $\Delta y_{1-2}$.

  $\begin{array}{c}\Delta y_{1-2}=\frac{0-{\left(120\text{ m/s}\right)}^2}{-2\left(9.80{\text{ m/s}}^2\right)}\\ =735\text{ m}\end{array}$

Apply equation (II) for the motion of the rocket from point $1$ to point $2$.

  $\begin{array}{c}{v}_{f_{1-2}}=v_{f_{0-1}}+\left(-g\right)t_{1-2}\\ =v_{f_{0-1}}-g{t}_{1-2}\end{array}$

Here, $t_{1-2}$ is the time taken by the rocket to reach point $2$ from point $1$.

Rewrite the above equation for $t_{1-2}$ .

  $t_{1-2}=\frac{v_{f_{1-2}}-v_{f_{0-1}}}{-g}$

Substitute $0$ for $v_{f_{1-2}}$, $120\text{ m/s}$ for $v_{f_{0-1}}$ and $9.80{\text{ m/s}}^2$ for $g$ in the above equation to find $t_{1-2}$.

  $\begin{array}{c}{t}_{1-2}=\frac{0-120\text{ m/s}}{-9.80{\text{ m/s}}^2}\\ =12.2\text{ s}\end{array}$

Apply equation (I) for the motion of the rocket from point $2$ to point $3$ .

  $\begin{array}{c}{v}_{f_{2-3}}^2=v_{f_{1-2}}^2+2\left(-g\right)\Delta y_{2-3}\\ =v_{f_{1-2}}^2-2g\Delta y_{2-3}\end{array}$        (III)

Here, $v_{f_{2-3}}$ is the speed with which the rocket strikes the ground and $\Delta y_{2-3}$ is the distance from point $2$ to point $3$.

Write the equation for $\Delta y_{2-3}$.

  $\Delta y_{2-3}=-\left(\Delta y_{0-1}+\Delta y_{1-2}\right)$

Substitute $1000\text{ m}$ for $\Delta y_{0-1}$ and $735\text{ m}$ for $\Delta y_{1-2}$ in the above equation to find $\Delta y_{2-3}$.

  $\begin{array}{c}\Delta y_{2-3}=-\left(1000\text{ m}+735\text{ m}\right)\\ =-1735\text{ m}\end{array}$

Substitute $0$ for $v_{f_{1-2}}$, $9.80{\text{ m/s}}^2$ for $g$ and $-1735\text{ m}$ for $\Delta y_{2-3}$ in equation (III) to find $v_{f_{2-3}}$.

  $\begin{array}{c}{v}_{f_{2-3}}^2=0-2\left(9.80{\text{ m/s}}^2\right)\left(-1735\text{ m}\right)\\ =34006{\text{ m}}^2{\text{/s}}^2\\ v_{f_{2-3}}=-184\text{ m/s}\end{array}$

The velocity is negative since it is directed downward.

Apply equation (II) for the motion of the rocket from point $2$ to point $3$.

  $\begin{array}{c}{v}_{f_{2-3}}=v_{f_{1-2}}+\left(-g\right)t_{2-3}\\ =v_{f_{1-2}}-g{t}_{2-3}\end{array}$

Here, $t_{2-3}$ is the time taken by the rocket to reach point $3$ from point $2$.

Rewrite the above equation for $t_{2-3}$ .

  $t_{2-3}=\frac{v_{f_{2-3}}-v_{f_{1-2}}}{-g}$

Substitute $-184\text{ m/s}$ for $v_{f_{2-3}}$, $0$ for $v_{f_{1-2}}$ and $9.80{\text{ m/s}}^2$ for $g$ in the above equation to find $t_{2-3}$.

  $\begin{array}{c}{t}_{2-3}=\frac{-184\text{ m/s}-0}{-9.80{\text{ m/s}}^2}\\ =18.8\text{ s}\end{array}$

Write the equation for the time interval the rocket is in motion above the ground.

  $t_{\text{tot}}=t_{0-1}+t_{1-2}+t_{2-3}$        (IV)

Here, $t_{\text{tot}}$ is the time interval the rocket is in motion above the ground.

Conclusion:

Substitute $10.0\text{ s}$ for $t_{0-1}$, $12.2\text{ s}$ for $t_{1-2}$ and $18.8\text{ s}$ for $t_{2-3}$ in equation (IV) to find $t_{\text{tot}}$ .

  $\begin{array}{l}{t}_{\text{tot}}=10.0\text{ s}+12.2\text{ s}+18.8\text{ s}\\ =41.0\text{ s}\end{array}$

Therefore, the time interval for which the rocket is in motion above the ground is $41.0\text{ s}$.

(b)

To determine

The maximum altitude of the rocket.

Answer to Problem 49P

The maximum altitude of the rocket is $1.73\text{ km}$.

Explanation of Solution

Write the equation for the maximum altitude.

  $\Delta y_{2-3}=\Delta y_{0-1}+\Delta y_{1-2}$

Conclusion:

Substitute $1000\text{ m}$ for $\Delta y_{0-1}$ and $735\text{ m}$ for $\Delta y_{1-2}$ in the above equation to find $\Delta y_{2-3}$.

  $\begin{array}{c}\Delta y_{2-3}=1000\text{ m}+735\text{ m}\\ =1735\text{ m}\frac{1\text{ km}}{1000\text{ m}}\\ =1.735\text{ km}\\ \approx \text{1}\text{.73 km}\end{array}$

Therefore, the maximum altitude of the rocket is $1.73\text{ km}$.

(c)

To determine

The velocity of the rocket just before it hits the ground.

Answer to Problem 49P

The velocity of the rocket just before it hits the ground is $-184\text{ m/s}$.

Explanation of Solution

Equation (III) gives the velocity with which the rocket strikes the ground.

Conclusion:

Substitute $0$ for $v_{f_{1-2}}$, $9.80{\text{ m/s}}^2$ for $g$ and $-1735\text{ m}$ for $\Delta y_{2-3}$ in equation (III) to find $v_{f_{2-3}}$.

  $\begin{array}{c}{v}_{f_{2-3}}^2=0-2\left(9.80{\text{ m/s}}^2\right)\left(-1735\text{ m}\right)\\ =34006{\text{ m}}^2{\text{/s}}^2\\ v_{f_{2-3}}=-184\text{ m/s}\end{array}$

Therefore, the velocity of the rocket just before it hits the ground is $-184\text{ m/s}$.