Chapter 3, Problem 18P

(a)

To determine

The time of his flight .

Answer to Problem 18P

The time of his flight is $\underset{\_}{0.852\text{s}}$.

Explanation of Solution

Write the expression for the vertical final velocity of the basketball player.

    $v_{yf}^2=v_{yi}^2+2a\left(y_f-y_i\right)$        (I)

Here, $v_{yf}$ is the vertical final velocity of the basketball player, $v_{yi}$ is the vertical initial velocity of the basketball player, $a_y$ is the vertical acceleration component of the player, $y_f$ is the final vertical position and $y_i$ is the initial vertical position.

Write the expression for the time of flight,

    $\begin{array}{c}{v}_{yf}=v_{if}+a_{y}t\\ t=\frac{v_{yf}-v_{if}}{a_y}\end{array}$        (II)

Conclusion:

For upward flight, $v_{yf}=0$.

Substitute $-9.8{\text{m/s}}^{\text{2}}$ for $a_y$, $0$ for $v_{yf}$, $1.85\text{m}$ for $y_f$ and $1.02\text{m}$ for $y_i$ in (I) to find $v_{yi}$

    $\begin{array}{c}0=v_{yi}^2+2\left(-9.8{\text{m/s}}^{\text{2}}\right)\left(1.85-1.02\right)\text{m}\\ {\text{v}}_{yi}\text{=4}\text{.03}\text{m/s}\end{array}$

For downward flight, $v_{yi}=0$.

Substitute $-9.8{\text{m/s}}^{\text{2}}$ for $a_y$, $0$ for $v_{yi}$, $0.90\text{m}$ for $y_f$ and $1.85\text{m}$ for $y_i$ in (I) to find $v_{yf}$

    $\begin{array}{c}{v}_{yf}^2=0+2\left(-9.8{\text{m/s}}^{\text{2}}\right)\left(0.900\text{m}-1.85\text{m}\right)\\ v_{yf}=-4.32\text{m/s}\end{array}$

Substitute $-9.8{\text{m/s}}^{\text{2}}$ for $a_y$, $4.03\text{m/s}$ for $v_{yi}$ and $-4.32\text{m/s}$ for $v_{yi}$ in (II) to find $t$

    $\begin{array}{c}t=\frac{\left(-4.32-4.03\right)\text{m/s}}{-9.8{\text{m/s}}^{\text{2}}}\\ =0.852\text{s}\end{array}$

Therefore, the time of his flight is $\underset{\_}{0.852\text{s}}$.

(b)

To determine

The horizontal velocity component of the basketball player at instant of take-off .

Answer to Problem 18P

The horizontal velocity component of the basketball player at instant of take-off is $\underset{\_}{3.29\text{m/s}}$.

Explanation of Solution

Write the total horizontal displacement of the basketball player,

    $x=v_{xi}t$        (III)

Here, $x$ is the total horizontal displacement of the basketball player and $v_{xi}$ is the initial horizontal velocity component of the basketball player.

Conclusion:

Substitute $2.80\text{m}$ for $x$ and $0.852\text{s}$ for $t$ in (III) to find $v_{xi}$ ,

    $\begin{array}{c}2.80\text{m}=v_{xi}\left(0.852\text{s}\right)\\ v_{xi}=3.29\text{m/s}\end{array}$

Therefore, the horizontal velocity component of the basketball player at instant of take-off is $\underset{\_}{3.29\text{m/s}}$.

(c)

To determine

The vertical velocity component of the basketball player at instant of take-off .

Answer to Problem 18P

The vertical velocity component of the basketball player at instant of take-off is $\underset{\_}{\text{4}\text{.03}\text{m/s}}$.

Explanation of Solution

From part (a), for upward flight,

Write the expression for the vertical final velocity of the basketball player.

    $v_{yf}^2=v_{yi}^2+2a\left(y_f-y_i\right)$        (I)

Here, $v_{yf}$ is the vertical final velocity of the basketball player, $v_{yi}$ is the vertical initial velocity of the basketball player, $a_y$ is the vertical acceleration component of the player, $y_f$ is the final vertical position and $y_i$ is the initial vertical position.

Conclusion:

For upward flight, $v_{yf}=0$.

Substitute $-9.8{\text{m/s}}^{\text{2}}$ for $a_y$, $0$ for $v_{yf}$, $1.85\text{m}$ for $y_f$ and $1.02\text{m}$ for $y_i$ in (I) to find $v_{yi}$

    $\begin{array}{c}0=v_{yi}^2+2\left(-9.8{\text{m/s}}^{\text{2}}\right)\left(1.85-1.02\right)\text{m}\\ {\text{v}}_{yi}\text{=4}\text{.03}\text{m/s}\end{array}$

Therefore, the vertical velocity component of the basketball player at instant of take-off is $\underset{\_}{\text{4}\text{.03}\text{m/s}}$.

(d)

To determine

The take-off angle of the basketball player .

Answer to Problem 18P

The take-off angle of the basketball player is $\underset{\_}{{50.8}^{\text{o}}}$.

Explanation of Solution

Write the take-off angle of the basketball player,

    $\theta ={\tan }^{-1}\frac{v_{yi}}{v_{xi}}$        (IV)

Here, $\theta$ is the take-off angle of the basketball player.

Conclusion:

Substitute $4.03\text{m/s}$ for $v_{yi}$ and $3.29\text{m/s}$ for $v_{xi}$ in (IV) to find $\theta$ ,

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{4.03\text{m/s}}{3.29\text{m/s}}\right)\\ ={50.8}^{\text{o}}\end{array}$

Therefore, the take-off angle of the basketball player is $\underset{\_}{{50.8}^{\text{o}}}$.

(e)

To determine

The flight time of the whitetail deer .

Answer to Problem 18P

The flight time of the whitetail deer is $\underset{\_}{1.12\text{s}}$.

Explanation of Solution

From part (a),

Write the expression for the vertical final velocity of the basketball player.

    $v_{yf}^2=v_{yi}^2+2a\left(y_f-y_i\right)$        (V)

Here, $v_{yf}$ is the vertical final velocity of the whitetail deer, $v_{yi}$ is the vertical initial velocity of the whitetail deer, $a_y$ is the vertical acceleration component of the whitetail deer, $y_f$ is the final vertical position of the whitetail deer and $y_i$ is the initial vertical position of the whitetail deer.

Write the expression for the flight time of the whitetail deer,

    $\begin{array}{c}{v}_{yf}=v_{if}+a_{y}t\\ t=\frac{v_{yf}-v_{if}}{a_y}\end{array}$        (VI)

Here, $t$ is the flight time of the whitetail deer.

Conclusion:

For upward flight, $v_{yf}=0$.

Substitute $-9.8{\text{m/s}}^{\text{2}}$ for $a_y$, $0$ for $v_{yf}$, $2.50\text{m}$ for $y_f$ and $1.20\text{m}$ for $y_i$ in (V) to find $v_{yi}$

    $\begin{array}{c}0=v_{yi}^2+2\left(-9.8{\text{m/s}}^{\text{2}}\right)\left(2.50-1.20\right)\text{m}\\ {\text{v}}_{yi}\text{=5}\text{.04}\text{m/s}\end{array}$

For downward flight, $v_{yi}=0$.

Substitute $-9.8{\text{m/s}}^{\text{2}}$ for $a_y$, $0$ for $v_{yi}$, $0.700\text{m}$ for $y_f$ and $2.50\text{m}$ for $y_i$ in (V) to find $v_{yf}$

    $\begin{array}{c}{v}_{yf}^2=0+2\left(-9.8{\text{m/s}}^{\text{2}}\right)\left(0.700\text{m}-2.50\text{m}\right)\\ v_{yf}=-5.94\text{m/s}\end{array}$

Substitute $-9.8{\text{m/s}}^{\text{2}}$ for $a_y$, $5.04\text{m/s}$ for $v_{yi}$ and $-5.94\text{m/s}$ for $v_{yf}$ in (VI) to find $t$

    $\begin{array}{c}t=\frac{-5.94\text{m/s}-5.04\text{m/s}}{-9.8{\text{m/s}}^{\text{2}}}\\ =1.12\text{s}\end{array}$

Therefore, the flight time of the whitetail deer is $\underset{\_}{1.12\text{s}}$.