Chapter 2, Problem 63AP

(a)

To determine

Time after which the two stones hit each other after the release of first stone.

Answer to Problem 63AP

The time after which the two stones hit each other after the release of the first stone is $\underset{\_}{3.00\text{s}}$.

Explanation of Solution

Let $y_i$ be the initial position and $y_f$ be the final position of the stone. Thus the displacement of the stone is equal to $y_f-y_i$.

Write the expression for the displacement of the first stone after the release.

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

Here, $y_f$ is the final position of the first stone, $y_i$ is the initial position of the stone, $v_{yi}$ is the initial velocity of the stone in the $y$ direction, $t$ is the time taken to fall into water, and $a_y$ is the acceleration.

Express equation (I) as a quadratic form.

    $-\frac{1}{2}{a}_y{t}^2-v_{yi}t+y_f-y_i=0$                                                                                       (II)

Write the general expression for a second order quadratic equation.

    $a{t}^2+bt+c=0$.                                                                                                        (III)

Write the expression for the solution of expression (III).

    $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$                                                                                           (IV)

Use expression (III) in (IV) and solve for $t$.

    $t=\frac{-v_{yi}\pm \sqrt{{\left(v_{yi}\right)}^2-4a_y{y}_f}}{2a_y}$                                                                                       (V)

Conclusion:

Substitute $-2.00\text{m/s}$ for $v_{yi}$, $-9.80{\text{m/s}}^2$ for $a_y$, and $-50.0\text{m}$ for $y_f$ in equation (III).

    $\begin{array}{c}-\frac{1}{2}\left(9.80{\text{m/s}}^2\right)t^2-\left(-2.00\text{m/s}\right)t-50.0\text{m}=0\\ \left(4.90{\text{m/s}}^2\right)t^2+\left(2.00\text{m/s}\right)t-50.0\text{m=0}\end{array}$                                                        (VI)

Compare expression (VII) with the general equation (III) and find the values of constants.

    $\begin{array}{c}{v}_{yi}=2.00\text{m/s}\\ a_y=4.90{\text{m/s}}^2\\ y_f=-50.0\text{m}\end{array}$

Substitute $2.00\text{m/s}$ for $v_{yi}$, $4.90{\text{m/s}}^2$ for $a_y$, and $-50.0\text{m}$ for $y_f$ in equation (V) to find $t$.

    $\begin{array}{c}t=\frac{-2.00\text{m/s}\pm \sqrt{{\left(2.00\text{m/s}\right)}^2-4\left(4.90{\text{m/s}}^2\right)\left(-50.0\text{m}\right)}}{2\left(4.90{\text{m/s}}^2\right)}\\ =3.00\text{s}\end{array}$

Therefore, the time after which the two stones hit each other is $\underset{\_}{3.00\text{s}}$.

(b)

To determine

The initial velocity of the stone if they are hitting each other simultaneously.

Answer to Problem 63AP

The initial velocity of the stone if they are hitting each other simultaneously is $\underset{\_}{-15.3\text{m/s}}$.

Explanation of Solution

The two stones are thrown vertically downward and they hit each other in a time interval of $1.00\text{s}$. The time of hitting of the two stones after the release of first stone is $3.00\text{s}$. Therefore the time of travel of second stone before hitting is equal to $3.00\text{s}-1.00\text{s}=2.00\text{s}$

    $t=2.00\text{s}$                                                                                                               (VII)

Use expression (I) to find $v_{yi}$.

    $\begin{array}{c}{y}_f-y_i=v_{yi}t+\frac{1}{2}{a}_y{t}^2\\ v_{yi}=\frac{\left(y_f-y_i\right)-\frac{1}{2}{a}_y{t}^2}{t}\end{array}$                                                                                (VIII)

Conclusion:

Substitute $-50.0\text{m}$ for $y_f$, $0\text{m}$ for $y_i$, $-9.80{\text{m/s}}^2$ for $a_y$, and $2.00\text{s}$ for $t$ in equation (VIII) to find $v_{yi}$.

    $\begin{array}{c}{v}_{yi}=\frac{\left(-50.0\text{m}\right)-\frac{1}{2}\left(-9.80{\text{m/s}}^2\right){\left(2.00\text{s}\right)}^2}{\left(2.00\text{s}\right)}\\ =-15.3\text{m/s}\end{array}$

Therefore, the initial velocity of the stone if they are hitting each other simultaneously is $\underset{\_}{-15.3\text{m/s}}$.

(c)

To determine

Speed of both stones when they hit the water.

Answer to Problem 63AP

Speed of first stone is $\underset{\_}{-31.4\text{m/s}}$ and the speed of second stone is $\underset{\_}{-34.8\text{m/s}}$.

Explanation of Solution

Write the expression for the final velocity of first stone.

    $v_{1f}=v_{1i}+a_1{t}_1$                                                                                                  (IX)

Here, $v_{1f}$ is the final velocity of first ball, $v_{1i}$ is the initial velocity of first ball, $a_1$ is the acceleration of first ball, and $t_1$ is the time of travel of first stone.

Write the expression for the final velocity of second stone.

    $v_{2f}=v_{2i}+a_2{t}_2$                                                                                                  (X)

Here, $v_{2f}$ is the final velocity of second ball, $v_{2i}$ is the initial velocity of second ball, $a_2$ is the acceleration of second stone, and $t_2$ is the time of travel of second stone.

Conclusion:

Substitute $-2.00\text{m/s}$ for $v_{1i}$, $-9.80{\text{m/s}}^2$ for $a_1$, and $3.00\text{s}$ for $t_1$ in equation (IX) to find $v_{1f}$.

    $\begin{array}{c}{v}_{1f}=-2.00\text{m/s}+\left(-9.80{\text{m/s}}^2\right)\left(3.00\text{s}\right)\\ =-31.4\text{m/s}\end{array}$

Substitute $-15.3\text{m/s}$ for $v_{2i}$, $-9.80{\text{m/s}}^2$ for $a_2$, and $2.00\text{s}$ for $t_2$ in equation (X) to find $v_{2f}$.

    $\begin{array}{c}{v}_{2f}=-15.3\text{m/s}+\left(-9.80{\text{m/s}}^2\right)\left(2.00\text{s}\right)\\ =-34.8\text{m/s}\end{array}$

Therefore, speed of first stone is $\underset{\_}{-31.4\text{m/s}}$ and the speed of second stone is $\underset{\_}{-34.8\text{m/s}}$.