Chapter 3, Problem 39AP

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 100. m/s. The rocket moves for 3.00 s along its initial line of motion with an acceleration of 30.0 m/s². At this time, its engines fail and the rocket proceeds to move as a projectile. 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 39AP

The maximum altitude reached by the rocket is $1.53\times {10}^3\text{m}$.

Explanation of Solution

The distance travelled is found from the kinematic equation,

$s=v_{0}t+\frac{1}{2}a{t}^2$

Here,

$v_0$ is the initial velocity

$a$ is the acceleration

$t$ is the time taken

Substitute $100\text{m/s}$ for $v_0$, $3.00\text{s}$ for $t$ and $30.0{\text{m/s}}^2$ for $a$.

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

The coordinates of the rocket at the end of the powered flight are,

$\begin{array}{l}{x}_1=s\cos \theta \\ y_1=s\sin \theta \end{array}$

Substitute $435\text{m}$ for $s$ and $53.0°$ for $\theta$.

$\begin{array}{c}{x}_1=\left(435\text{m}\right)\cos 53.0°\\ =262\text{m}\\ y_1=\left(435\text{m}\right)\sin 53.0°\\ =347\text{m}\end{array}$

The speed of the rocket at the end of the powered flight is,

$v_1=v_0+at$

Substitute $100\text{m/s}$ for $v_0$, $3.00\text{s}$ for $t$ and $30.0{\text{m/s}}^2$ for $a$.

$\begin{array}{c}{v}_1=100\text{m/s}+\left(30.0{\text{m/s}}^2\right)\left(3.00\text{s}\right)\\ =190\text{m/s}\end{array}$

The components of the initial velocity of the rocket are,

$\begin{array}{l}{v}_{0x}=v_1\cos \theta \\ v_{0y}=v_1\sin \theta \end{array}$

Substitute $190\text{m/s}$ for $v_1$ and $53.0°$ for $\theta$.

$\begin{array}{c}{v}_{0x}=\left(190\text{m}\right)\cos 53.0°\\ =114\text{m/s}\\ v_{0y}=\left(190\text{m}\right)\sin 53.0°\\ =152\text{m/s}\end{array}$

The rise time during the free fall phase is,

$t_{\text{rise}}=\frac{v_y-v_{0y}}{a_y}$

Here,

$v_y$ is the vertical velocity

$v_{0y}$ is the vertical component of the initial velocity

$a_y$ is the vertical acceleration

$\Delta y$ is the vertical displacement

Substitute $0\text{m/s}$ for $v_y$, $152\text{m/s}$ for $v_{0y}$ and $-9.80{\text{m/s}}^2$ for $a_y$.

$\begin{array}{c}{t}_{\text{rise}}=\frac{0\text{m/s}-152\text{m/s}}{-9.80{\text{m/s}}^2}\\ =15.5\text{s}\end{array}$

The vertical displacement during this time is,

$\Delta y=\left(\frac{v_y+v_{0y}}{2}\right)t_{\text{rise}}$

Substitute $0\text{m/s}$ for $v_y$, $152\text{m/s}$ for $v_{0y}$ and $15.5\text{s}$ for $t_{\text{rise}}$.

$\begin{array}{c}\Delta y=\left(\frac{0\text{m/s}+152\text{m/s}}{2}\right)15.5\text{s}\\ =1180\text{m}\end{array}$

The maximum altitude reached is,

$H=y_1+\Delta y$

Substitute $347\text{m}$ for $y_1$ and $1180\text{m}$ for $\Delta y$.

$\begin{array}{c}H=347\text{m}+1180\text{m}\\ =1.53\times {10}^3\text{m}\end{array}$

Conclusion:

Thus, the maximum altitude reached by the rocket is $1.53\times {10}^3\text{m}$.

To determine

The total time of flight.

Answer to Problem 39AP

The total time of flight is $36.3\text{s}$.

Explanation of Solution

The time taken for the rocket to fall a distance of $1.53\times {10}^3\text{m}$ is,

$t_{\text{fall}}=\sqrt{\frac{2\left(\Delta y\right)}{a_y}}$

Substitute $-1.53\times {10}^3\text{m}$ for $\Delta y$ and $-9.80{\text{m/s}}^2$ for $a_y$.

$\begin{array}{c}{t}_{\text{fall}}=\sqrt{\frac{2\left(-1.53\times {10}^3\text{m}\right)}{-9.80{\text{m/s}}^2}}\\ =17.7\text{s}\end{array}$

The total time of flight is,

$t=t_{\text{powered}}+t_{\text{rise}}+t_{\text{fall}}$

Substitute $3.00\text{s}$ for $t_{\text{powered}}$, $15.5\text{s}$ for $t_{\text{rise}}$ and $17.7\text{s}$ for $t_{\text{fall}}$.

$\begin{array}{c}t=3.00\text{s}+15.5\text{s}+17.7\text{s}\\ =36.2\text{s}\end{array}$

Conclusion:

The total time of flight is $36.3\text{s}$.

To determine

The horizontal range.

Answer to Problem 39AP

The horizontal range is $4.04\times {10}^3\text{m}$.

Explanation of Solution

The rocket is in free fall for,

$t=t_{\text{rise}}+t_{\text{fall}}$

Substitute $15.5\text{s}$ for $t_{\text{rise}}$ and $17.7\text{s}$ for $t_{\text{fall}}$.

$\begin{array}{c}{t}_2=15.5\text{s}+17.7\text{s}\\ =33.2\text{s}\end{array}$

The horizontal displacement during this time is,

$\Delta x=v_{0x}{t}_2$

Substitute $114\text{m/s}$ for $v_{0x}$ and $33.2\text{s}$ for $t_2$.

$\begin{array}{c}\Delta x=\left(114\text{m/s}\right)\left(33.2\text{s}\right)\\ =3.78\times {10}^3\text{m}\end{array}$

The full horizontal range is,

$R=x_1+\Delta x$

Substitute $262\text{m}$ for $x_1$ and $3.78\times {10}^3\text{m}$ for $\Delta x$.

$\begin{array}{c}R=262\text{m}+3.78\times {10}^3\text{m}\\ =4.04\times {10}^3\text{m}\end{array}$

Conclusion:

Thus, the horizontal range is $4.04\times {10}^3\text{m}$.