Chapter 6, Problem 137P

To determine

The final speeds of the three rocks.

Answer to Problem 137P

The final speeds of the three rocks is $\underset{\_}{20\text{m}/\text{s}}$.

Explanation of Solution

Given that the initial speed of the rock is $6.0\text{m}/\text{s}$, and the vertical distance travelled by the rock is $18.0\text{m}$.

Considering rock A, B, and C are dropped from same vertical height $h$, and with same initial speed.

From conservation of energy principle, energy associated with the rock A before dropping and after dropping are equal.

    $\frac{1}{2}{m}_{\text{A}}{v}_i^2+m_{\text{A}}gh=\frac{1}{2}{m}_{\text{A}}{v}_{\text{A}}^2$

Here, $m_{\text{A}}$ is the mass of rock A, $v_i$ is the initial velocity of the rock A, $g$ is the acceleration due to gravity of the rock, $h$ is the vertical distance travelled by the rock A, and $v_{\text{A}}$ is the final speed of the rock A.

Rearrange the above equation to find $v_{\text{A}}$.

    $\begin{array}{c}{v}_i^2+2gh=v_{\text{A}}^2\\ v_{\text{A}}^2=v_i^2+2gh\\ v_{\text{A}}=\sqrt{v_i^2+2gh}\end{array}$        (I)

From conservation of energy principle, energy associated with the rock B before dropping and after dropping are equal.

    $\frac{1}{2}{m}_{\text{B}}{v}_i^2+m_{\text{B}}gh=\frac{1}{2}{m}_{\text{B}}{v}_{\text{B}}^2$

Here, $m_{\text{B}}$ is the mass of rock B, $v_i$ is the initial velocity of the rock B, $g$ is the acceleration due to gravity of the rock, $h$ is the vertical distance travelled by the rock B, and $v_{\text{B}}$ is the final speed of the rock B.

Rearrange the above equation to find $v_{\text{B}}$.

    $\begin{array}{c}{v}_i^2+2gh=v_{\text{B}}^2\\ v_{\text{B}}^2=v_i^2+2gh\\ v_{\text{B}}=\sqrt{v_i^2+2gh}\end{array}$        (II)

From conservation of energy principle, energy associated with the rock C before dropping and after dropping are equal.

    $\frac{1}{2}{m}_{\text{C}}{v}_i^2+m_{\text{C}}gh=\frac{1}{2}{m}_{\text{C}}{v}_{\text{C}}^2$

Here, $m_{\text{C}}$ is the mass of rock C, $v_i$ is the initial velocity of the rock C, $g$ is the acceleration due to gravity of the rock, $h$ is the vertical distance travelled by the rock C, and $v_{\text{C}}$ is the final speed of the rock C.

Rearrange the above equation to find $v_{\text{C}}$.

    $\begin{array}{c}{v}_i^2+2gh=v_{\text{C}}^2\\ v_{\text{C}}^2=v_i^2+2gh\\ v_{\text{C}}=\sqrt{v_i^2+2gh}\end{array}$        (II)

From equation (I), (II) and (III), it is clear the final velocity of the rock A, B, and C are equal $\left(v_{\text{A}}=v_{\text{B}}=v_{\text{C}}\right)$ since all the parameters in equation (I), (II) and (III) are having same value.

Conclusion:

Substitute $6.0\text{m}/\text{s}$ for $v_{\text{i}}$, $9.8\text{m}/{\text{s}}^2$ for $g$, and $18.0\text{m}$ for $h$ in equation (I), to find $v_{\text{A}}$.

    $\begin{array}{c}{v}_{\text{A}}=\sqrt{{\left(6.0\text{m}/\text{s}\right)}^2+2\left(9.8\text{m}/{\text{s}}^2\right)\left(18.0\text{m}\right)}\\ =\sqrt{388.8}\text{m}/\text{s}\\ =19.7\text{m}/\text{s}\\ \approx 20\text{m}/\text{s}\end{array}$

Since $\left(v_{\text{A}}=v_{\text{B}}=v_{\text{C}}\right)$ the final velocity of the rock A, B, and C is $20\text{m}/\text{s}$.

Therefore, the final speeds of the three rocks is $\underset{\_}{20\text{m}/\text{s}}$.