Chapter 4, Problem 65AP

(a)

To determine

The maximum altitude.

Answer to Problem 65AP

The maximum altitude for this projectile motion is $1.52\text{km}$.

Explanation of Solution

Draw the below figure for this condition as.

This is the case of projectile motion. Before the engine fails, car has acceleration up to distance from point $o$ to point $A$. This acceleration has two components in vertical and horizontal direction.

Write the expression for vertical acceleration as.

    $a_y=a\sin \theta$                                                                                                    (I)

Here, $a_y$ is vertical acceleration, $a$ is acceleration and $\theta$ is projection angle.

Write the expression for horizontal acceleration as.

    $a_x=a\cos \theta$                                                                                                  (II)

Here, $a_x$ is horizontal acceleration.

Write the expression for initial velocity along vertical as.

    $v_{yi}=v\sin \theta$                                                                                                 (III)

Here, $v_{yi}$ is initial velocity along vertical and $v$ is iniytial velocity.

Write the expression for initial velocity along horizontal as.

    $v_{xi}=v\cos \theta$                                                                                                 (IV)

Here, $v_{xi}$ is initial velocity along horizontal.

This projectile motion is divided into three part. Calculate the distances traveled during each phase of the motion.

Case:1

Refer to above figure at point $O$.

Write the expression for final velocity along vertical at point $O$ as.

    $v_{yf}=v_{yi}+a_{y}t$                                                                                               (V)

Here, $v_{yf}$ is final velocity along vertical, $v_{yi}$ is initial velocity along vertical and $t$ is time.

Write the expression for final velocity along horizontal as.

    $v_{xf}=v_{xi}+a_{x}t$                                                                                              (VI)

Here, $v_{xf}$ is final velocity along horizontal and $v_{xi}$ is initial velocity along horizontal.

Write the expression for vertical displacement as.

    $\Delta y_1=v_{yi}t+\frac{1}{2}{a}_y{t}^2$                                                                                       (VII)

Here, $\Delta y_1$ is vertical displacement.

Write the expression for horizontal displacement as.

    $\Delta x_1=v_{xi}t+\frac{1}{2}{a}_x{t}^2$                                                                                      (VIII)

Here, $\Delta x_1$ is horizontal displacement.

Case:2

Refer to the drawn figure, at point $A$, horizontal acceleration is zero as speed is same for the entire path. The initial velocity along horiontal is equal to the final velocity along horizontal. For this case, final velocity along vertical is also zero.

Rearrange equation (V) in terms of time $t$ as.

    $t=\frac{v_{yf}-v_{yi}}{a_y}$                                                                                                (IX)

Write the expression for horizontal distance cover by the engine as.

    $\Delta x_2=v_{xf}t$                                                                                                      (X)

Write the expression for vertical distance cover by the engine as.

    $\Delta y_2=v_{yf}t-\frac{1}{2}a{t}^2$                                                                                         (XI)

Write the expression for maximum altitude as.

    $\Delta y=\Delta y_1+\Delta y_2$                                                                                           (XII)

Here, $\Delta y$ is maximum altitude.

Case:3

Refer to above figure, at point $B$, horizontal acceleration is zero as speed is same for the entire path. The initial velocity along horiontal is equal to the final velocity along horizontal. For this case, final velocity along vertical is also zero.

Use third equation of motion to calculate the final velocity along verical at point $B$.

Write the expression for equation of motion at point $B$ as.

    $v_{yf}=\sqrt{2a\Delta y}$                                                                                             (XIII)

Conclusion:

Substitute $30.0{\text{ m/s}}^2$ for $a$ and $53.0°$ for $\theta$ in equation (I)

    $\begin{array}{c}{a}_y=30.0{\text{ m/s}}^2\left(\sin 53.0°\right)\\ =24.0{\text{ m/s}}^2\end{array}$

Substitute $30.0{\text{ m/s}}^2$ for $a$ and $53.0°$ for $\theta$ in equation (II).

    $\begin{array}{c}{a}_x=30.0{\text{ m/s}}^2\left(\cos 53.0°\right)\\ =18.1{\text{ m/s}}^2\end{array}$

Substitute $100.0\text{ m/s}$ for $v$ and $53.0°$ for $\theta$ in equation (III).

    $\begin{array}{c}{v}_{yi}=100.0\text{ m/s}\left(\sin 53.0°\right)\\ =79.9\text{ m/s}\end{array}$

Substitute $100.0\text{ m/s}$ for $v$ and $53.0°$ for $\theta$ in equation (IV).

    $\begin{array}{c}{v}_{xi}=100.0\text{ m/s}\left(\cos 53.0°\right)\\ =60.2\text{ m/s}\end{array}$

Substitute $79.9\text{ m/s}$ for $v_{yi}$, $24.0{\text{ m/s}}^2$ for $a_y$ and $3.00\text{ s}$ for $t$ in equation (V).

    $\begin{array}{c}{v}_{yf}=79.9\text{ m/s}+\left(24.0{\text{ m/s}}^2\right)\left(3.00\text{ s}\right)\\ =79.9\text{ m/s}+72.0\text{ m/s}\\ \text{=152}\text{.0 m/s}\end{array}$

Substitute $60.2\text{ m/s}$ for $v_{xi}$, $18.1{\text{ m/s}}^2$ for $a_x$ and $3.00\text{ s}$ for $t$ in equation (VI).

    $\begin{array}{c}{v}_{xf}=60.2\text{ m/s}+\left(18.1{\text{ m/s}}^2\right)\left(3.00\text{ s}\right)\\ =60.2\text{ m/s}+54.3\text{ m/s}\\ \text{=114 m/s}\end{array}$

Substitute $79.9\text{ m/s}$ for $v_{yi}$ and $24.0{\text{ m/s}}^2$ for $a_y$ in equation (VII).

    $\begin{array}{c}\Delta y_1=79.9\text{ m/s}\left(3.00\text{ s}\right)+\frac{1}{2}\left(24.0{\text{ m/s}}^2\right){\left(3.00\text{ s}\right)}^2\\ =239.7\text{ m}+108.0\text{ m}\\ \text{=347}\text{.7 m}\end{array}$

Substitute $60.2\text{ m/s}$ for $v_{xi}$, $18.1{\text{ m/s}}^2$ for $a_x$ and $3.00\text{ s}$ for $t$ in equation (VIII).

    $\begin{array}{c}\Delta x_1=60.2\text{ m/s}\left(3.00\text{ s}\right)+\frac{1}{2}\left(18.1{\text{ m/s}}^2\right){\left(3.00\text{ s}\right)}^2\\ =180.6\text{ m}+81.45\text{ m}\\ \text{=262}\text{m}\end{array}$

Substitute $\text{0}$ for $v_{xf}$ $152.0\text{ m/s}$ for $v_{yi}$ and $-9.80{\text{ m/s}}^2$ for $a_y$ in equation (IX).

  $\begin{array}{c}t=\frac{\text{0}-152.0\text{ m/s}}{-9.80{\text{ m/s}}^2}\\ =15.5\text{ s}\end{array}$

Substitute $114.5\text{ m/s}$ for $v_{xf}$ and $15.5\text{ s}$ for $t$ in equation (X).

    $\begin{array}{c}\Delta x_2=114.5\text{ m/s}\left(15.5\text{ s}\right)\\ =1774.75\text{ m}\\ \text{=1}\text{.77 km}\end{array}$

Substitute $152\text{ m/s}$ for $v_{yf}$, $15.5\text{ s}$ for $t$ and $9.80{\text{ m/s}}^2$ for $a$ in equation (XI)

    $\begin{array}{c}\Delta y_2=152\text{ m/s}\left(15.5\text{ s}\right)-\frac{1}{2}\left(9.80{\text{ m/s}}^2\right){\left(\text{15}\text{.5 s}\right)}^2\\ =2356.0\text{ m}-1177.225\text{ m}\\ \text{=1178}\text{.775 m}\\ \text{=1}\text{.17 km}\end{array}$

Substitute $\text{347}\text{.7 m}$ for $\Delta y_1$ and $1.17\text{ km}$ for $\Delta y_2$ in equation (XII)

    $\begin{array}{c}\Delta y=\text{347}\text{.7 m}+1.17\text{ km}\\ \text{=1517}\text{.7 m}\\ \text{=1}\text{.52 km}\end{array}$

Thus, the maximum altitude for this projectile motion is $1.52\text{km}$.

(b)

To determine

The total time of fight.

Answer to Problem 65AP

The total time of flight is $36.1\text{ s}$.

Explanation of Solution

The acceleration is under gravity. The final velocity along vertical is downward direction at point $B$.

Write the expression for total time of flight as.

    $t_3=\frac{-v_{yf}}{-g}$                                                                                                  (XIV)

Here, $t_3$ is time for projectile path $DB$.

Write the expression for total flight time as.

    $\Delta t=t_1+t_2+t_3$                                                                                            (XV)

Here, $\Delta t$ is total flight time, $t_1$ is time for projectile path $OA$, $t_2$ is time for projectile path $AD$ and $t_3$ is time for projectile path $DB$.

Conclusion:

Substitute $-1.52\text{ km}$ for $\Delta y$ and $v_{yf}$$-9.80{\text{ m/s}}^2$ for $a$ in equation (XIII)

    $\begin{array}{c}{v}_{yf}=\sqrt{2\left(-9.80{\text{ m/s}}^2\right)\left(-1.52\text{ km}\right)}\\ =\sqrt{\left(19.6{\text{ m/s}}^2\right)\left(1.52\text{ km}\left(\frac{1000\text{ m}}{1\text{ km}}\right)\right)}\\ =172.60\text{ m/s}\\ \text{=173 m/s}\end{array}$

Substitute $173\text{ m/s}$ for $v_{yf}$ and $9.80{\text{ m/s}}^2$ for $g$ in equation (XIV)

    $\begin{array}{c}t=\frac{-173\text{ m/s}}{-9.80{\text{ m/s}}^2}\\ =17.6\text{ s}\end{array}$

Substitute $3.00\text{ s}$ for $t_1$, $15.5\text{ s}$ for $t_2$ and $17.65\text{ s}$ for $t_3$ in equation (XV)

    $\begin{array}{c}\Delta t=3.00\text{ s}+15.5\text{ s}+17.6\text{ s}\\ \text{=36}\text{.1 s}\end{array}$

Thus, the total time of flight is $36.1\text{ s}$.

(c)

To determine

The horizontal range.

Answer to Problem 65AP

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

Explanation of Solution

Refer to above figure, the horizontal range can be calculated by the sum of distance $OC$ and $CB$.

Write the expression for horizontal distance cover by the engine as.

    $\Delta x_3=v_{xf}{t}_3$                                                                                               (XVI)

Here, $\Delta x_3$ is horizontal distance covered by engine at the time of free fall.

Write the expression for horizontal range as.

    $\Delta x=\Delta x_1+\Delta x_2+\Delta x_3$                                                                              (XVII)

Here $\Delta x$ is net horizontal range

Conclusion:

Substitute $114.5\text{ m/s}$ for $v_{xf}$ and $17.65\text{ s}$ for $t_3$ in equation (XVI)

    $\begin{array}{c}\Delta x_3=114.5\text{ m/s}\left(17.65\text{ s}\right)\\ =2.006\times {10}^3\text{ m}\end{array}$

Substitute $262\text{ m}$ for $\Delta x_1$, $1.77\times {10}^3\text{ m}$ for $\Delta x_2$ and $2.006\times {10}^3\text{ m}$ for $\Delta x_3$ in equation (XVII)

    $\begin{array}{c}\Delta x=262\text{ m}+1.77\times {10}^3\text{ m}+2.006\times {10}^3\text{ m}\\ =4.04\times {10}^3\text{ m}\end{array}$

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