Chapter 4, Problem 32PQ

Some cats can be trained to jump from one location to another and perform other tricks. Kit the cat is going to jump through a hoop. He begins on a wicker cabinet at a height of 1.750 m above the floor and jumps through the center of a vertical hoop, reaching a peak height 3.125 m above the floor. a. With what initial velocity did Kit leave the cabinet if the hoop is at a horizontal distance of 1.544 m from the cabinet? b. If Kit lands on a bed at a horizontal distance of 3.587 m from the cabinet, how high above the ground is the bed?

To determine

The initial velocity of Kit.

Answer to Problem 32PQ

The initial velocity of Kit is $\overrightarrow{v}=\left(2.92\widehat{i}+5.18\widehat{j}\right)\text{m}/\text{s}$.

Explanation of Solution

Kit is the cat that is trained to jump from one location to another through the center of a vertically placed hoop. Kit is at a height of $y_i=1.750\text{ m}$ and after jumping through the hoop is reached a peak height of $y_f=3.125\text{ m}$ above the floor. If the horizontal distance of the hoop is $x_f-x_i=\Delta x=1.544\text{ m}$, the initial velocity of Kit be.

Write the formula for the vertical final velocity [Kinematic equation for constant acceleration]

    $v_{yf}^2=v_{yi}^2+2a_y\Delta y$                                                                                                 (I)

Here, $v_{yi}$ is the initial vertical velocity, $v_{yf}$ is the final vertical velocity, $a_y$ is the vertical acceleration, $\Delta y$ is the vertical displacement.

Write the formula for the vertical position

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

Here, $y_f$ is the final position in y axis, $y_i$ is the initial position in y axis,

Write the formula for the horizontal position

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

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

Kit jumped from initial position $1.750\text{ m}$ to $3.125\text{ m}$, therefore, the total vertical displacement of Kit is $3.125\text{ m}-1.750\text{ m}=1.375\text{ m}$. The final vertical velocity of Kit after reached a peak height of $y_f=3.125\text{ m}$ above the floor is $v_{yf}=0$ as Kit is standing still. The vertical acceleration is due to gravity pulling the Kit in direction downward, $a_y=-9.81\text{m}/{\text{s}}^{\text{2}}$.

Substituting $1.375\text{ m}$ for $\Delta y$, $-9.81\text{m}/{\text{s}}^{\text{2}}$ for $a_y$ in equation (I) to find the value of $v_{yi}$

$\begin{array}{l}0=v_{yi}^2+2\left(-9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(1.375\text{ m}\right)\\ v_{yi}=5.19\text{m}/\text{s}\end{array}$

The initial vertical velocity of Kit is $5.19\text{m}/\text{s}$.

Substituting $5.19\text{m}/\text{s}$ for $v_{yi}$, $-9.81\text{m}/{\text{s}}^{\text{2}}$ for $a_y$ in equation (II) to find the value of $t$

$\begin{array}{c}0=5.19\text{m}/\text{s}-\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)t\\ t=0.529\text{s}\end{array}$

The time taken for Kit to reach the hoop is $0.529\text{s}$.

Substitute $1.544\text{ m}$ for $\Delta x$ and $0.529\text{s}$ for $t$ in equation (III) to find the value of $v_{xi}$

$\begin{array}{c}\Delta x=v_{xi}t\\ 1.544\text{ m}=v_{xi}\left(0.529\text{s}\right)\\ v_{xi}=2.92\text{m}/\text{s}\end{array}$

The initial horizontal velocity of Kit is $2.92\text{m}/\text{s}$ and the initial vertical velocity of Kit is $5.19\text{m}/\text{s}$. Therefore, in component form, the initial velocity of Kit is $\overrightarrow{v}=\left(2.92\widehat{i}+5.18\widehat{j}\right)\text{m}/\text{s}$.

To determine

How high above the ground is the bed.

Answer to Problem 32PQ

The bed is $0.72\text{ m}$ above the ground.

Explanation of Solution

The Kit lands on a bed at a horizontal distance of $3.587\text{ m}$ away from its initial position $y_i=1.750\text{ m}$.

Substitute $1.544\text{ m}$ for $\Delta x$ and $2.92\text{m}/\text{s}$ for $v_{xi}$ in equation (III) to find the value of $t$

$\begin{array}{c}3.587\text{ m}=2.92\text{ s}\times t\\ t=1.23\text{ s}\end{array}$

The time taken for the Kit to land on the bed is $1.23\text{ s}$.

Substituting $5.19\text{m}/\text{s}$ for $v_{yi}$, $1.23\text{ s}$ for $t$, $1.750\text{ m}$ for $y_i$ and $-9.81\text{m}/{\text{s}}^{\text{2}}$ for $a_y$ in equation (II) to find the value of $y_f$

$\begin{array}{l}{y}_f=1.750\text{ m}+\left(5.19\text{ s}\right)\left(1.23\text{ s}\right)-\frac{1}{2}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right){\left(1.23\text{ s}\right)}^2\\ y_f=\left(1.750+6.39-7.42\right)\text{ m}\\ y_f=0.72\text{ m}\end{array}$

Thus, the bed is $0.72\text{ m}$ above the ground.