Chapter 4, Problem 20P

(a)

To determine

The maximum height reached by the rock.

Answer to Problem 20P

The maximum height reached by the rock is $77.05\text{1 m}$

Explanation of Solution

Write the equation of motion.

  $v^2-u^2=2as$        (I)

Here, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, and $s$ is the height.

Conclusion:

Substitute $0$ for $v$, ${0.634}^2$ for $u$, $\text{-9}{\text{.8m/s}}^{\text{2}}$ for $a$ in expression (I)

    $\begin{array}{c}{v}^2-u^2=2as\\ 0-{6.34}^2=2\times (-9.8)\times s\\ s=2.051\text{m}\end{array}$

Therefore the maximum height of the rock reach is $2.051\text{m}$ plus the height of the bridge,

  $75+2.051=77.05\text{1 m}$

(b)

To determine

The time required by the rock to reach the maximum height.

Answer to Problem 20P

The time required by the rock to reach the maximum height is $0.647\sec$

Explanation of Solution

Write the equation of motion.

  $v=u+at$        (I)

Here, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, and $t$ is the time interval.

Conclusion:

Substitute $0$ for $v$, ${0.634}^2$ for $u$, $\text{-9}{\text{.8m/s}}^{\text{2}}$ for $a$ in expression (I)

  $\begin{array}{c}0\text{ = }6.34-9.8t\\ t\text{ = }0.647\sec \end{array}$

The time required by the rock to reach the maximum height is $0.647\sec$

(c)

To determine

The place where the rock lands.

Answer to Problem 20P

The place where the rock lands is $62.7\text{m}$

Explanation of Solution

Write the equation of motion.

  $s=ut+\frac{1}{2}a{t}^2$        (I)

Here, $u$ is the initial velocity, $a$ is the acceleration, $s$ is the height, and $t$ is the time interval.

Conclusion:

Substitute $0$ for $u$, $77.051$ for $u$, $\text{-9}{\text{.8m/s}}^{\text{2}}$ for $a$ in expression (I)

  $\begin{array}{c}s=ut+\frac{1}{2}a{t}^2\\ 77.051=0+0.5\times 9.8kt{t}^2\\ t=3.965s\end{array}$

Total time taken to reach ground is $3.965+0.647=\text{ 4}\text{.612 s}$

Therefore the total horizontal displacement is,

  $x_{lands} =v_{0x}{t}_{lands}$        (II)

Substitute $\text{15m/s}$ for $v_{0x}$, $\text{4}\text{.61s}$ for $t_{lands}$ in expression (II)

  $x_{lands}=\left(\text{15m/s}\right)\left(\text{cos25°}\right)\left(\text{4}\text{.61s}\right)=63\text{ m}$

The place where the rock lands is $62.7\text{m}$

(d)

To determine

The time at which the rock lands.

Answer to Problem 20P

The time at which the rock lands is $\text{ 4}\text{.612 s}$

Explanation of Solution

Write the equation of motion.

  $s=ut+\frac{1}{2}a{t}^2$        (I)

Here, $u$ is the initial velocity, $a$ is the acceleration, $s$ is the height, and $t$ is the time interval.

Conclusion:

Substitute $0$ for $u$, $77.051$ for $u$, $\text{-9}{\text{.8m/s}}^{\text{2}}$ for $a$ in expression (I)

  $\begin{array}{c}s=ut+\frac{1}{2}a{t}^2\\ 77.051=0+0.5\times 9.8kt{t}^2\\ t=3.965s\end{array}$

Total time taken to reach ground is $3.965+0.647=4.612s$

The time at which the rock lands is $\text{ 4}\text{.612 s}$

(e)

To determine

The velocity of the rock just before it lands.

Answer to Problem 20P

The magnitude of the velocity $4\text{1}\text{.167 m/s}$ and $-70.72°$.

Explanation of Solution

Write the equation of motion.

  $v=u+at$        (I)

Here, $u$ is the initial velocity, $a$ is the acceleration, $s$ is the height, and $t$ is the time interval.

Conclusion:

Substitute $0$ for $u$, $3.965$ for $t$, $\text{9}{\text{.8m/s}}^{\text{2}}$ for $a$ in expression (I)

  $\begin{array}{c}{v}_y=u+at\\ v_y= 0+9.8\times 3.965\\ =38.\text{857 m/s}\end{array}$

And the velocity along ${\text{v}}_x= 13.59\text{5 m/s}$

The magnitude of the velocity,

  $\begin{array}{c}\left|\text{v}\right|= \sqrt{{\left(13.59\text{5 m/s}\right)}^2+{\left(38.857\right)}^2}\\ = 41.16\text{7 m/s}\end{array}$

Write the direction will then be,

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{v_y}{v_x}\right)\\ \theta ={\tan }^{-1}\left(\frac{38.9\text{ m/s}}{13.6\text{ m/s}}\right)\\ \theta =-70.72°\end{array}$