Chapter 4, Problem 4.65AP

A catapult launches a rocket at an angle of 53.0° above the horizontal with an initial speed of 100 m/s. The rocket engine immediately starts a burn, and for 3.00 s the rocket moves along its initial line of motion with an acceleration of 30.0 m/s². Then its engine fails, and the rocket proceeds to move in free fall. Find (a) the maximum altitude reached by the rocket, (b) its total time of flight, and (c) its horizontal range.

To determine

The maximum altitude reached by the rocket.

Answer to Problem 4.65AP

The maximum altitude reached by the rocket is $1.52\text{km}$.

Explanation of Solution

Given info: The initial speed of the rocket is $100\text{m}/\text{s}$ with an angle $53.0°$ above the horizontal, the time for which the rocket runs along its initial line of motion is $3.00\text{s}$ and the acceleration is $30.0{\text{m}/\text{s}}^2$.

The formula to calculate the vertical height reached by the rocket is,

$h=\left({\left(v_0\right)}_{\text{y}}\sin \theta \right)t+\frac{1}{2}{a}_{\text{y}}{t}^2$

Here,

$h$ is the vertical height.

${\left(v_0\right)}_{\text{y}}$ is the vertical component of the initial velocity.

$t$ is the time taken by the rocket.

$a_{\text{y}}$ is the vertical component acceleration of the rocket.

$\theta$ is the angle made by the rocket with the horizontal.

Substitute $100\text{m}/\text{s}$ for ${\left(v_0\right)}_{\text{y}}$, $53.0°$ for $\theta$, $30.0{\text{m}/\text{s}}^2$ for $a_{\text{y}}$ and $3.00\text{s}$ for $t$ in the above equation.

$\begin{array}{c}h=\left(\left(100\text{m}/\text{s}\right)\sin \left(53.0°\right)\right)\left(3.00\text{s}\right)+\frac{1}{2}\left(30.0{\text{m}/\text{s}}^2\right){\left(3.00\text{s}\right)}^2\\ =\left(\left(100\text{m}/\text{s}\right)\sin \left(53.0°\right)\right)\left(3.00\text{s}\right)+\frac{1}{2}\left(30.0{\text{m}/\text{s}}^2\right){\left(3.00\text{s}\right)}^2\\ =239.59\text{m}+135\text{m}\\ =\text{374}\text{.59}\text{m}\end{array}$

Thus, the vertical height of rocket is $\text{374}\text{.59}\text{m}$.

The speed of the rocket after the failure is,

$v_0={\left(v_0\right)}_{\text{y}}+a_{\text{y}}t$

Substitute $100\text{m}/\text{s}$ for ${\left(v_0\right)}_{\text{y}}$, $30.0{\text{m}/\text{s}}^2$ for $a_{\text{y}}$ and $3.00\text{s}$ for $t$ in the above equation.

$\begin{array}{c}{v}_0=100\text{m}/\text{s}+\left(30.0{\text{m}/\text{s}}^2\right)\left(3.00\text{s}\right)\\ =190\text{m}/\text{s}\end{array}$

The height reached by the rocket after the engine failure is,

$h_{\text{f}}=\frac{v^2-v_0^2{\sin }^2\theta }{-2g}$

Here,

$v$ is the final velocity of the rocket.

$g$ is the acceleration due to gravity.

$h_{\text{f}}$ is the height reached by the rocket after the engine failure.

Substitute $0$ for $v$, $190\text{m}/\text{s}$ for $v_0$, $53.0°$ for $\theta$ and $10.0{\text{m}/\text{s}}^2$ for $g$ in the above equation.

$\begin{array}{c}{h}_{\text{f}}=\frac{0^2-{\left(190\text{m}/\text{s}\right)}^2{\sin }^2\left(53.0°\right)}{-2\left(10.0{\text{m}/\text{s}}^2\right)}\\ =1151.26\text{m}\end{array}$

The maximum height reached by the rocket is,

$h_{\max }=h_{\text{f}}+h$

Substitute $1151.26\text{m}$ for $h_{\text{f}}$ and $\text{374}\text{.59}\text{m}$ for $h$ in the above equation.

$\begin{array}{c}{h}_{\max }=1151.26\text{m}+\text{374}\text{.59}\text{m}\\ =1525.8\text{m}\left(\frac{{10}^{-3}\text{km}}{1\text{m}}\right)\\ =1.52\text{km}\end{array}$

Conclusion:

Therefore, the maximum altitude reached by the rocket is $1.52\text{km}$.

To determine

The total time of flight of the rocket.

Answer to Problem 4.65AP

The total time of flight of the rocket is $\text{36}.1\text{s}$.

Explanation of Solution

Given info: The initial speed of the rocket is $100\text{m}/\text{s}$ with an angle $53.0°$ above the horizontal, the time for which the rocket runs along its initial line of motion is $3.00\text{s}$ and the acceleration is $30.0{\text{m}/\text{s}}^2$.

The vertical height reached by the rocket at the time of free fall is,

$h=\left(v_0\sin \theta \right)t^{\prime }+\frac{1}{2}g{t^{\prime }}^2$

Here,

$t^{\prime }$ is the time of flight of rocket after free fall.

Substitute $190\text{m}/\text{s}$ for $v_0$, $53.0°$ for $\theta$, $9.8{\text{m}/\text{s}}^2$ for $g$ and $\text{374}\text{.59}\text{m}$ for $h$ in the above equation.

$\begin{array}{c}-\text{374}\text{.59}\text{m}=\left(190\text{m}/\text{s}\right)\sin \left(53.0°\right)t^{\prime }+\frac{1}{2}\left(9.8{\text{m}/\text{s}}^2\right){t^{\prime }}^2\\ -\text{374}\text{.59}\text{m}=151.7\text{m}/\text{s}{t}^{\prime }-\left(4.9{\text{m}/\text{s}}^2\right){t^{\prime }}^2\\ t^{\prime }=33.10\text{s,-2}\text{.14}\text{s}\\ t^{\prime }=33.10\text{s}\end{array}$

The total time of the flight of the rocket is,

$T=t+t^{\prime }$

Substitute $33.10\text{s}$ for $t^{\prime }$ and $3.00\text{s}$ for $t$ in the above equation.

$\begin{array}{c}T=3.00\text{s}+33.10\text{s}\\ =\text{36}.1\text{s}\end{array}$

Conclusion:

Therefore, the total time of flight of the rocket is $\text{36}.1\text{s}$.

To determine

The horizontal range of the rocket.

Answer to Problem 4.65AP

The horizontal range of the rocket is $\text{4}\text{.05}\text{km}$.

Explanation of Solution

Given info: The initial speed of the rocket is $100\text{m}/\text{s}$ with an angle $53.0°$ above the horizontal, the time for which the rocket runs along its initial line of motion is $3.00\text{s}$ and the acceleration is $30.0{\text{m}/\text{s}}^2$.

The formula to calculate the displacement of the rocket along the initial line of motion is,

$h^{\prime }={\left(v_0\right)}_{\text{y}}t+\frac{1}{2}{a}_{\text{y}}{t}^2$

Here,

$h^{\prime }$ is the displacement of the rocket along the initial line of motion.

Substitute $100\text{m}/\text{s}$ for ${\left(v_0\right)}_{\text{y}}$, $30.0{\text{m}/\text{s}}^2$ for $a_{\text{y}}$ and $3.00\text{s}$ for $t$ in the above equation.

$\begin{array}{c}{h}^{\prime }=\left(100\text{m}/\text{s}\right)\left(3.00\text{s}\right)+\frac{1}{2}\left(30.0{\text{m}/\text{s}}^2\right){\left(3.00\text{s}\right)}^2\\ =300.0\text{m}+135.5\text{m}\\ =\text{435}\text{.0}\text{m}\end{array}$

Thus, the displacement of the rocket along the initial line of motion is $\text{435}\text{.0}\text{m}$.

The horizontal range of the rocket during constant acceleration is,

$R^{\prime }=h^{\prime }\cos \theta$

Here,

$R^{\prime }$ is the horizontal range of the rocket during constant acceleration.

Substitute $435.0\text{m}$ for $h^{\prime }$ and $53.0°$ for $\theta$ in the above equation.

$\begin{array}{c}{R}^{\prime }=\left(435.0\text{m}\right)\cos \left(53.0°\right)\\ =261.8\text{m}\end{array}$

The horizontal range of the rocket during free fall is calculated as,

$R^{″}=\left(v_0\cos \theta \right)T$

Here,

$R^{″}$ is the horizontal range of the rocket during free fall.

Substitute $190\text{m}/\text{s}$ for $v_0$, $53.0°$ for $\theta$ and $33.1\text{s}$ for $T$ in the above equation.

$\begin{array}{c}{R}^{″}=\left(190\text{m}/\text{s}\right)\cos \left(53.0°\right)\left(33.1\text{s}\right)\\ =3784.81\text{m}\end{array}$

The total horizontal range of the rocket is,

$R=R^{\prime }+R^{″}$

Substitute $261.8\text{m}$ for $R^{\prime }$ and $3784.81\text{m}$ for $R^{″}$ in the above equation.

$\begin{array}{c}R=261.8\text{m}+3784.81\text{m}\\ =4046.61\text{m}\left(\frac{{10}^{-3}\text{km}}{1\text{m}}\right)\\ =4.0466\text{km}\\ \approx \text{4}\text{.05}\text{km}\end{array}$

Conclusion:

Therefore, the horizontal range of the rocket is $\text{4}\text{.05}\text{km}$.