Chapter 4, Problem 33PQ

Dock diving is a great form of athletic competition for dogs of all shapes and sizes (Fig. P4.33). Sheba, the American Pit Bull Terrier, runs and jumps off the dock with an initial speed of 9.02 m/s at an angle of 25° with respect to the surface of the water. a. If Sheba begins at a height of 0.84 m above the surface of the water, through what horizontal distance does she travel before hitting the surface of the water? b. Write an expression for the velocity of Sheba, in component form, the instant before she hits the water. c. Determine the peak height above the water reached by Sheba during her jump.

FIGURE P4.33

To determine

The horizontal distance travelled by Sheba before hitting the surface of the water.

Answer to Problem 33PQ

The horizontal distance travelled by Sheba before hitting the surface of the water is $7.8\text{ m}$.

Explanation of Solution

Sheba, the dog, is dock diving in an athletic competition for dogs and jumps off the dock with an initial speed of $v_i=9.02\text{m}/\text{s}$ at an angle $25°$ with respect to the surface of the water. The initial height of Sheba is $y_i=0.84\text{ m}$ above the surface of the water.

Write the formula for the vertical position

    $y_f=y_i+v_{yi}t+\frac{1}{2}{a}_y{t}^2$                                                                                            (I)

Here, $y_f$ is the final position in y axis, $y_i$ is the initial position in y axis, $a_y$ is the vertical acceleration, $t$ is the time and $v_{yi}$ is the initial vertical velocity.

Write the formula for the horizontal position

    $x_f=x_i+v_{xi}t$                                                                                                       (II)

Here, $x_f$ is the final position in $x$ axis, $x_i$ is the initial position in $x$ axis and $v_{xi}$ is the horizontal velocity

Sheba jumped from initial position $y_i=0.84\text{ m}$. The initial vertical velocity of Sheba is

$\begin{array}{c}{v}_{yf}=v_i\sin \theta \\ =\left(9.02\text{m}/\text{s}\right)\sin \left(25°\right)\\ =3.81\text{m}/\text{s}\end{array}$.

The vertical acceleration is due to gravity pulling the Sheba in direction downward, $a_y=-9.81\text{m}/{\text{s}}^{\text{2}}$.

Substituting $0.84\text{ m}$ for $y_i$, $3.81\text{m}/\text{s}$ for $v_{yi}$ and $-9.81\text{m}/{\text{s}}^{\text{2}}$ for $a_y$ in equation (I) to find the value of $v_{yi}$

$0=0.84\text{ m}+\left(3.81\text{m}/\text{s}\right)t-\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)t$

Solving the quadratic equation, the valuable root is $t=0.96\text{ s}$.

The time taken for Sheba to reach the water surface is $0.96\text{ s}$. The horizontal velocity of Sheba is

 $\begin{array}{c}{v}_{xi}=v_i\cos \theta \\ =\left(9.02\text{m}/\text{s}\right)\cos \left(25°\right)\\ =8.17\text{m}/\text{s}\end{array}$

Substitute $8.17\text{m}/\text{s}$ for $v_{xi}$ and $0.96\text{ s}$ for $t$ in equation (III) to find the value of $\Delta x$

$\begin{array}{c}\Delta x=v_{xi}t\\ \Delta x=\left(8.17\text{m}/\text{s}\right)\left(0.96\text{ s}\right)\\ \Delta x=7.8\text{ m}\end{array}$

Thus, the horizontal distance travelled by Sheba before hitting the surface of the water is $7.8\text{ m}$.

To determine

Write the velocity of Sheba at the instant before she hits the water in component form.

Answer to Problem 33PQ

The velocity of Sheba at the instant before she hits the water in component form is $\overrightarrow{v}=\left(8.2\widehat{i}-5.6\widehat{j}\right)\text{m}/\text{s}$.

Explanation of Solution

Write the formula for the velocity vector equation for the projectile motion

    $\overrightarrow{v}=\left(v_i\cos \theta \right)\widehat{i}+\left(v_i\sin \theta -gt\right)\widehat{j}$                                                                              (III)

Here, $v_i$ is the initial velocity, $\theta$ is the angle, $g$ is the acceleration due to gravity and $\overrightarrow{v}$ is the velocity vector.

The horizontal velocity component is $v_x=v_i\cos \theta =\left(9.02\text{m}/{\text{s}}^{\text{2}}\right)\cos \left(25°\right)=8.2\text{m}/\text{s}$.

The vertical velocity component is $v_y=v_i\sin \theta -gt=\left(9.02\text{m}/{\text{s}}^{\text{2}}\right)\sin \left(25°\right)-\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(0.96\text{ s}\right)=-5.6\text{m}/\text{s}$.

Substituting in equation (III) to find the value of $\overrightarrow{v}$

$\overrightarrow{v}=\left(8.2\widehat{i}-5.6\widehat{j}\right)\text{m}/\text{s}$

Thus, the velocity of Sheba at the instant before she hits the water in component form is $\overrightarrow{v}=\left(8.2\widehat{i}-5.6\widehat{j}\right)\text{m}/\text{s}$.

To determine

The peak height above the water surface reached by Sheba during her jump.

Answer to Problem 33PQ

The peak height above the water surface reached by Sheba during her jump is $1.58\text{ m}$.

Explanation of Solution

Write the formula for the vertical velocity component for the projectile motion

    $v_y=v_{yf}-gt$                                                                              (IV)

The initial vertical velocity of Sheba is $v_{yf}=v_i\sin \theta =\left(9.02\text{m}/\text{s}\right)\sin \left(25°\right)=3.81\text{m}/\text{s}$. The vertical velocity of Sheba at maximum height is zero, $v_y=0$.

Substituting $3.81\text{m}/\text{s}$ for $v_{yf}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (I) to find the value of $t$

$\begin{array}{l}0=3.81\text{m}/\text{s}-\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)t\\ t=\frac{3.81}{9.8}\text{ s}\\ t=0.39\text{ s}\end{array}$

The time taken for Sheba to reach the peak height is $0.39\text{ s}$.

Substituting $0.84\text{ m}$ for $y_i$, $3.81\text{m}/\text{s}$ for $v_{yi}$, $0.39\text{ s}$ for $t$  and $-9.81\text{m}/{\text{s}}^{\text{2}}$ for $a_y$ in equation (I) to find the value of $y_f$

$\begin{array}{l}{y}_f=0.84\text{ m}+\left(3.81\text{m}/\text{s}\right)\left(0.39\text{ s}\right)-\frac{1}{2}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right){\left(0.39\text{ s}\right)}^2\\ y_f=1.58\text{ m}\end{array}$

Thus, the peak height above the water surface reached by Sheba during her jump is $1.58\text{ m}$.