Chapter 4, Problem 175P

To determine

The horizontal distance before the driveway at which J should drop the paper.

Answer to Problem 175P

The horizontal distance before the driveway at which J should drop the paper is $36\text{ m}$ .

Explanation of Solution

Write the equation to find the time taken for the newspaper to hit the ground.

$\Delta y=vt+\frac{1}{2}g{t}^2$

Here, $\Delta y$ is the height from which the newspaper is dropped, $v$ is the initial velocity of the paper, $g$ is the acceleration due to gravity and $t$ is the time taken for the newspaper to hit the ground

The initial velocity of the newspaper is zero.

Substitute $0$ for $v$ in the above equation and rewrite the equation for $t$ .

$\begin{array}{c}\Delta y=0+\frac{1}{2}g{t}^2\\ =\frac{1}{2}g{t}^2\\ t=\sqrt{\frac{2\Delta y}{g}}\end{array}$ (I)

Write the equation for the horizontal distance travelled by the newspaper as it travels through the air.

$\Delta x_1=v_{i}t$ (II)

Here, $\Delta x_1$ is the horizontal distance travelled by the newspaper as it travels through the air and $v_i$ is the speed of the newspaper

Write the Newton’s second law.

$\Sigma F_x=ma$ (III)

Here, $\Sigma F_x$ is the net force in $x$ direction, $m$ is the mass of the object and $a$ is the acceleration of the object

Write the equation for the net force in $x$ direction.

$\Sigma F_x=-f_k$ (IV)

Here, $f_k$ is the force of kinetic friction

Write the equation for force of kinetic friction.

$f_k={\mu }_{k}mg$

Here, ${\mu }_k$ is the coefficient of kinetic friction and $g$ is the acceleration due to gravity

Put the above equation in equation (IV).

$\Sigma F_x=-{\mu }_{k}mg$ (V)

Equate equations (III) and (V).

$\begin{array}{c}ma=-{\mu }_{k}mg\\ a=-{\mu }_{k}g\end{array}$ (VI)

Write the equation of motion to find the horizontal distance travelled by the newspaper after it hits the ground.

$v_f^2-v_i^2=2a\Delta x_2$

Here, $v_f$ is the final velocity of the newspaper and $\Delta x_2$ is the horizontal distance travelled by the newspaper after it hits the ground

Put equation (VI) in the above equation.

$v_f^2-v_i^2=-2{\mu }_{k}g\Delta x_2$

The final speed of the newspaper is zero.

Substitute $0$ for $v_f$ in the above equation and rewrite it for $\Delta x_2$ .

$\begin{array}{c}0-v_i^2=-2{\mu }_{k}g\Delta x_2\\ v_i^2=2{\mu }_{k}g\Delta x_2\\ \Delta x_2=\frac{v_i^2}{2{\mu }_{k}g}\end{array}$ (VII)

Write the equation for the horizontal distance before the driveway at which J should drop the paper.

$d=\Delta x_1+\Delta x_2$ (VIII)

Here, $d$ is the horizontal distance before the driveway at which J should drop the paper

Conclusion:

Given that the speed of the vehicle is $15\text{ m/s}$ , the height of the window is $1.00\text{ m}$ and the coefficient of kinetic friction is $0.40$ . The value of acceleration due to gravity is $9.8{\text{ m/s}}^2$ .

Substitute $1.00\text{ m}$ for $\Delta y$ and $9.8{\text{ m/s}}^2$ for $g$ in equation (I) to find $t$ .

$\begin{array}{c}t=\sqrt{\frac{2\left(1.00\text{ m}\right)}{9.8{\text{ m/s}}^2}}\\ =0.45\text{ s}\end{array}$

Substitute $15\text{ m/s}$ for $v_i$ and $0.45\text{ s}$ for $t$ in equation (II) to find $\Delta x_1$ .

$\begin{array}{c}\Delta x_1=\left(15\text{ m/s}\right)\left(0.45\text{ s}\right)\\ =6.8\text{ m}\end{array}$

Substitute $15\text{ m/s}$ for $v_i$ , $0.40$ for ${\mu }_k$ and $9.8{\text{ m/s}}^2$ for $g$ in equation (VII) to find $\Delta x_2$ .

$\begin{array}{c}\Delta x_2=\frac{{\left(15\text{ m/s}\right)}^2}{2\left(0.40\right)\left(9.8{\text{ m/s}}^2\right)}\\ =29\text{ m}\end{array}$

Substitute $6.8\text{ m}$ for $\Delta x_1$ and $29\text{ m}$ for $\Delta x_2$ in equation (VIII) to find $d$ .

$\begin{array}{c}d=6.8\text{ m}+29\text{ m}\\ =35.8\text{ m}\\ \approx \text{36 m}\end{array}$

Therefore, the horizontal distance before the driveway at which J should drop the paper is $36\text{ m}$ .