Tag Archives: College Physics Complete Solution Manual

College Physics by Openstax Chapter 3 Problem 3

Find the north and east components of the displacement for the hikers shown in Figure 3.50.

Solution:

Refer to Figure 3-3-A for the north and east components of the displacement s of the hikers.

Considering the right triangle formed. The north component is computed as

\begin{align*} \text{s}_{\text{north}} & = \left( 5.00 \ \text{km} \right)\sin 40^\circ \\ \text{s}_{\text{north}} & = 3.21 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}

Using the same right triangle, the east component is computed as follows.

\begin{align*} \text{s}_{\text{east}} & = \left( 5.00 \ \text{km} \right)\cos 40^\circ \\ \text{s}_{\text{east}} & = 3.83 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}

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College Physics by Openstax Chapter 2 Problem 57

A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s after being released, and (c) the time before it hits the ground.

Solution:

Part A

Consider Figure A.

We are interested in two positions. Position 1 is where the coin is dropped. At this position, the coin is 300 m above the ground, the time is 0 s, and the velocity is 10.0 m/s upward.

Position 2 is the highest point of the coin reaches. At this position, the velocity is equal to 0 m/s.

Position 1 is the initial position and position 2 is the final position. Solve for the value of y₂.

\begin{align*} \left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 +2a \Delta y \\ \Delta y & = \frac{\left( v_{y_2} \right)^2 - \left( v_{y_1} \right)^2}{2a} \\ y_2 - y_1 & = \frac{\left( v_{y_2} \right)^2 - \left( v_{y_1} \right)^2}{2a} \\ y_2& = \frac{\left( v_{y_2} \right)^2 - \left( v_{y_1} \right)^2}{2a} +y_1 \\ y_2 & = \frac{\left( 0 \ \text{m/s} \right)^2-\left( 10.0\ \text{m/s} \right)^2}{2\left( -9.81 \ \text{m/s}^2 \right)}+300\ \text{m} \\ y_2 & =305 \ \text{m} \ \qquad {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}

$\therefore$ The maximum height reached by the coin is about 305 meters from the ground.

Part B

We do not know the position 4 seconds after the coin has been released, the answer can be above or below the initial point. We can actually use one of the kinematical equations to solve for the final position given the time. Here, the initial position is the point of release and the final position is the point of interest at 4.00 seconds after release.

\begin{align*} \Delta y & = v_{y_1}t+\frac{1}{2}at^2 \\ y_2 - y_1 & = v_{y_1}t+\frac{1}{2}at^2 \\ y_2 & = y_1 + v_{y_1}t+\frac{1}{2}at^2 \\ y_2 & = 300\ \text{m} +\left( 10\ \text{m/s} \right)\left( 4.00\ \text{s} \right)+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( 4.00\ \text{s} \right)^2\\ y_2 & = 261.52\ \text{m} \\ y_2 & = 262\ \text{m}\ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\ \end{align*}

$\therefore$ The coin is at a height of 262 meters above the ground 4.00 seconds after release. That is, the coin is already dropping and it is already below the release point.

Solving for the velocity 4.00 seconds after release considering the same initial and final position.

\begin{align*} v_{y_2} & = v_{y_1}+at \\ v_{y_2} & = 10\ \text{m/s} + \left( -9.81\ \text{m/s}^2 \right)\left( 4.00 \ \text{s} \right) \\ v_{y_2} & = -29.2\ \text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

$\therefore$ The coin has a velocity of 29.2 m/s directed downward 4.00 seconds after it is released. This confirms that the coin is indeed moving downwards at this point.

Part C

Considering figure C, we have two positions. Position 1 is the point of release 300 m above the ground with a velocity of 10 m/s upward. This is time 0 s.

The second position is at the ground where y=0 m. We are interested at the time in this position.

Considering position 1 as the initial position and position 2 as the final position.

\begin{align*} \Delta y & = v_{y_1} \Delta t+\frac{1}{2}a\left( \Delta t \right)^2 \\ y_2-y_1 & = v_{y_1}t+\frac{1}{2}at^2 \\ 0\ \text{m}-300\ \text{m} & = \left( 10 \ \text{m/s} \right)t+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)t^2 \\ -300 & = 10t-4.905t^2 \\ 4.905t^2-10t-300 & = 0 \\ \end{align*}

Solve for the value of t using the quadratic formula with a=4.905, b=-10, and c=-300.

\begin{align*} t & = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ t & = \frac{-\left( -10 \right) \pm \sqrt{\left( 10 \right)^2-4\left( 4.905 \right)\left( -300 \right)}}{2\left( 4.905 \right)}\\ t & = 8.91 \ \text{s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

$\therefore$ The time is about 8.91 seconds before the coin hits the ground.

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College Physics by Openstax Chapter 2 Problem 56

A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms (8.00×10⁻⁵ s) . (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

Solution:

Part A

For this part, we shall consider Figure A.

We will be considering the two positions as shown. The first position is when the ball is dropped from a height of 1.50 meters. For this position, we know that y₁=1.50 m, t₁=0 s, and v_y1=0 m/s.

Position 2 is immediately after the ball hits the floor. For this position, we do not the time elapse and the velocity but we know that the height is zero. That is y₂=0 m.

Position 1 is the initial position and position 2 is the final position. Solving for v_y2, we have

\begin{align*} \left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\ \left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1 \right) \\ v_{y_2} & = \sqrt{\left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1 \right)} \\ v_{y_2} & = - \sqrt{\left( 0 \right)^2+2\left( -9.81\ \text{m/s}^2 \right)\left( 0\ \text{m} - 1.50\ \text{m} \right)} \\ v_{y_2} & = - 5.42 \ \text{m/s} \qquad {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

$\therefore$ The steel ball has a velocity of about 5.42 m/s directed downward when it strikes the floor.

Part B

Figure B shows the two positions we are interested in to solve for this part.

Position 1 is at the floor immediately just after the ball hits it. At this initial position, we have y1=0 m, and t1= 0 s. We do not know the velocity at this point.

At position 2, the ball bounced back to its second peak. We know that at the peak of a free falling body, the velocity is zero. So, for this final position, we have y₂=1.45 m, and v_y2=0 m/s. We do not know the time at this position.

Position 1 is the initial position while position 2 is the final position. Solving for the initial velocity we have

\begin{align*} \left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\ \left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1 \right) \\ \left( v_{y_1} \right)^2 & = \left( v_{y_2} \right)^2 - 2 a \left( y_2 - y_1 \right) \\ v_{y_1} & = \sqrt{\left( v_{y_2} \right)^2 - 2 a \left( y_2 - y_1 \right)} \\ v_{y_1} & = \sqrt{\left( 0 \ \text{m/s} \right)^2 -2\left( -9.81 \ \text{m/s}^2 \right) \left( 1.45 \ \text{m} - 0 \ \text{m} \right) }\\ v_{y_1} & = 5.33 \ \text{m/s} \qquad {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

$\therefore$ The steel ball has a velocity of about 5.33 m/s directed upward immediately after it leaves the floor.

Part C

From our answers in Part A and Part B, we have a change in velocity from -5.42 m/s to 5.33 m/s. So, in this case the initial velocity is v₁=-5.42 m/s and the final velocity is v₂=5.33 m/s. We can compute for the acceleration:

\begin{align*} a & = \frac{\Delta v}{\Delta t} \\ a & = \frac{v_2-v_1}{\Delta t} \\ a & = \frac{5.33 \ \text{m/s} - \left( -5.42 \ \text{m/s} \right)}{8.00 \times 10^{-5} \ \text{s}} \\ a & = 134,375 \ \text{m/s}^2 \\ a & = 1.34 \times 10^{5} \text{m/s}^2 \ \qquad {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}

Part D

The period of compression happens when the ball has a velocity of -5.42 m/s until it reaches 0 m/s. We shall solve for the change in displacement for this two given velocities. The initial velocity is -5.42 m/s and the final velocity is 0 m/s. The acceleration during this period is the one solved in Part C, a=1.34×10⁵ m/s².

\begin{align*} \left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\ \Delta y & = \frac{\left( v_{y_2} \right)^2 - \left( v_{y_1} \right)^2}{2 a} \\ \Delta y & = \frac{\left( 0 \ \text{m/s} \right)^2 - \left( -5.42 \ \text{m/s} \right)^2}{2\left( 1.34 \times 10^5 \ \text{m/s}^2 \right)}\\ \Delta y & = -0.000110 \ \text{m} \\ \Delta y & = -1.10 \times 10^{-4} \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}

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College Physics by Openstax Chapter 2 Problem 55

Suppose you drop a rock into a dark well and, using precision equipment, you measure the time for the sound of a splash to return.
(a) Neglecting the time required for sound to travel up the well, calculate the distance to the water if the sound returns in 2.0000 s.
(b) Now calculate the distance taking into account the time for sound to travel up the well. The speed of sound is 332.00 m/s in this well.

Solution:

Part A

Consider Figure A.

We shall consider two points for our solution. First, position 1 is the top of the well. In this position, we know that y₁=0, t₁=0 and v_y1=0.

Position 2 is located at the top of the water table where the rock will meet the water. Since we neglect the time for the sound to travel from position 2 to position 1, we can say that t₂=2.0000 s, the time of the rock to reach this position.

Solving for the value of y₂ will determine the distance between the two positions.

\begin{align*} \Delta y & = v_{y_1}t+\frac{1}{2}at^2 \\ y_2-y_1 & = v_{y_1}t+\frac{1}{2}at^2 \\ y_2 & = y_1 +v_{y_1}t+\frac{1}{2}at^2 \\ y_2 & = 0+0+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( 2.0000\ \text{s} \right)^2 \\ y_2 & = -19.6\ \text{m} \qquad {\color{DarkOrange} \left( \text{Answer }\right)} \end{align*}

Position 2 is 19.6 meters measured downward from position 1.

$\therefore$ The distance to the water is about 19.6 meters.

Part B

For this case, the 2.0000 seconds that is given includes the time that the rock travels from position 1 to position 2, t_r, and the time that the sound travels from position 2 to position 1, t_s.

\begin{align*} t_r+t_s & =2.0000\ \text{s} \\ t_s & = 2.0000\ \text{s}-t_r \end{align*}

Considering the motion of the rock from position 1 to position 2.

\begin{align*} \Delta y & = v_{y_1}t_r+\frac{1}{2}a\left( t_r \right)^2 \\ y_2-y_1 & = v_{y_1}t_r+\frac{1}{2}a\left( t_r \right)^2 \\ y_2 & = y_1 +v_{y_1}t_r+\frac{1}{2}a\left( t_r \right)^2 \\ y_2 & = 0+0+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( t_r \right)^2 \\ y_2 & = -4.905\left( t_r \right)^2 \qquad {\color{Blue} \text{Equation 1}}\\ \end{align*}

Now, let us consider the motion of the sound from position 2 to position 1. Sound is assumed to have a constant velocity of 322.00 m/s.

\begin{align*} \Delta y & = v_s \times t_s \\ y_1-y_2 & =\left( 322.00\ \text{m/s} \right)\left( t_s \right) \\ 0-y_2 & =\left( 322.00 \right)\left( 2.0000-t_r \right) \\ y_2 & = -322.00\left( 2.0000-t_r \right) \qquad {\color{Blue} \text{Equation 2}} \end{align*}

So, we have two equations from the two motions. We can solve the equations simultaneously.

\begin{align*} -4.905 \left( t_r \right)^2 & = -322.00\left( 2.0000-t_r \right) \\ 4.905 \left( t_r \right)^2 & =322.00\left( 2.0000-t_r \right) \\ 4.905 \left( t_r \right)^2 & = 644.00-322.00t_r \\ 4.905\left( t_r \right)^2 + 322.00t_r-644.00 & = 0 \\ \end{align*}

We can solve the quadratic formula using the quadratic equation.

\begin{align*} t_r & = \frac{-b \pm\sqrt{b^2-4ac}}{2a} \\ t_r & = \frac{-322.00\pm\sqrt{\left( 322.00 \right)^2-4\left( 4.905 \right)\left( -644.00 \right)}}{2\left( 4.905 \right)}\\ t_r & =1.9425 \ \text{s} \end{align*}

Now that we have solved for the value of t_r, we can use this to solve for y₂ using either Equation 1 or Equation 2. We will use equation 1.

\begin{align*} y_2 & = -4.905\left( t_r \right)^2 \\ y_2 & = -4.905 \left( 1.9425 \right)^2 \\ y_2 & =-18.5 \ \text{m} \qquad {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Position 2 is about 18.5 meters below position 1.

$\therefore$ In this case, the distance between the two positions is 18.5 meters.

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College Physics by Openstax Chapter 2 Problem 54

A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 0.312 s to go past the window. What was the ball’s initial velocity? Hint: First consider only the distance along the window, and solve for the ball’s velocity at the bottom of the window. Next, consider only the distance from the ground to the bottom of the window, and solve for the initial velocity using the velocity at the bottom of the window as the final velocity.

Solution:

First, we have the position 1 where the motion starts. Here, we know that y<sub>1</sub>=0, t<sub>1</sub>=0, and v<sub>y1</sub>=0.

Position 2 is at the bottom of the window. We know that it is 7.50 meters from where the motion started. So we have y<sub>2</sub>=7.50 meters. We do not know the time and velocity at this point.

Then we have position 3 at the top of the window where the overall height is 9.50 meters, y<sub>3</sub>=9.50. We also do not know the velocity and time elapsed in this position. — *Figure A*

Consider Figure A. We shall be considering the three positions shown.

First, we have position 1 where the motion starts. Here, we know that y₁=0 and t₁=0, but we do not know v_y1.

Position 2 is at the bottom of the window. We know that it is 7.50 meters from where the motion started. So we have y₂=7.50 meters. We do not know the time and velocity at this point.

Then we have position 3 at the top of the window where the overall height is 9.50 meters, y₃=9.50. We also do not know the velocity and time elapsed in this position.

Consider positions 2 and 3. The initial position in this case is at position 2 and the final position is at position 3. We know that the difference of time between this two positions is 0.312 seconds. We can say that

t_3 =t_2+0.312 \ \text{s} \\ t_3-t_2 = 0.312\ \text{s}

Using the same 2 positions still, we have

\begin{align*} y_3 & = y_2 + v_{y_2} \Delta t+\frac{1}{2}a\left( \Delta t \right)^2 \\ 9.50\ \text{m} & = 7.50\ \text{m} + v_{y_2} \left( t_3-t_2 \right)+\frac{1}{2}a\left( t_3-t_2 \right)^2 \\ 9.50\ \text{m}-7.50\ \text{m} & = v_{y_2}\left( 0.312\ \text{s} \right)+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( 0.312\ \text{s} \right)^2\\ 2.00\ \text{m} & = 0.312\ \text{s} \left( v_{y_2} \right)-0.4775\ \text{m} \\ 0.312\ \text{s} \left( v_{y_2} \right) & = 2.00\ \text{m}+0.4775\ \text{m} \\ 0.312\ \text{s} \left( v_{y_2} \right) & = 2.4775\ \text{m} \\ v_{y_2}& =\frac{2.4775\ \text{m}}{0.312\ \text{s}} \\ v_{y_2}& = 7.94\ \text{m/s} \end{align*}

We have computed the velocity of the ball at the bottom of the window.

Next, we shall consider positions 1 and 2. In this consideration, position 1 will be considered the initial position while position2 is the final position.

\begin{align*} \left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 +2a \Delta y \\ \left( 7.94\ \text{m/s} \right)^2 & = \left( v_{y_1} \right)^2 + 2\left( -9.81 \ \text{m/s}^2 \right)\left( 7.50\ \text{m}-0 \right)\\ \left( v_{y_1} \right)^2 & = \left( 7.94\ \text{m/s} \right)^2- 2\left( -9.81 \ \text{m/s}^2 \right)\left( 7.50\ \text{m}-0 \right)\\ v_{y_1} & = + \sqrt{ \left( 7.94\ \text{m/s} \right)^2- 2\left( -9.81 \ \text{m/s}^2 \right)\left( 7.50\ \text{m}-0 \right)} \\ v_{y_1} & = + 14.5\ \text{m/s}\qquad {\color{DarkOrange} \left( \text{Answer} \right) } \end{align*}

$\therefore$ The ball’s initial velocity is about 14.5 m/s upward.

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College Physics by Openstax Chapter 2 Problem 52

An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.

Solution:

Consider Figure 1. The object was dropped from a height of 75.0 m. At the start of motion, the velocity is zero, $v_{oy}=0$ .

The object traveled for a period of time t for the whole 75.0 m distance to the ground.

Part A

We are solving for the distance traveled by the object for the first 1 second. So, we have

\begin{align*} \Delta y & = v_{oy} t + \frac{1}{2}at^2\\ \Delta y & = \left( 0 \ \text{m/s} \right)\left( 1 \ \text{s} \right)+\frac{1}{2}\left( -9.81 \ \text{m/s}^2 \right)\left( 1 \ \text{s} \right)^2 \\ \Delta y & = 0 -4.905 \ \text{m} \\ \Delta y & = -4.91 \ \text{m} \\ |\Delta y| & =4.91 \ \text{m} \end{align*}

The negative sign of Δy indicates that the direction of the displacement is downward. Since we are looking for the scalar value of the distance, the answer is 4.91 m.

Part B

So we now consider the two positions of the object as shown in the figure to the right. The initial height of the object is 75.0 m above the ground, and the initial velocity is 0.

At the ground, we know that the position of the object is 0 m above the ground, but we do not know the time and velocity. Therefore, to determine the velocity of the object at this point, we proceed as follows:

\begin{align*} \left( v_2 \right)^2 & =\left( v_1 \right)^2+2a \Delta y \\ \left( v_2 \right)^2 & =\left( v_1 \right)^2+2a \left( y_2-y_1 \right) \\ v_2 & = \pm \sqrt{\left( v_1 \right)^2+2a \left( y_2-y_1 \right)}\\ v_2 & = \pm \sqrt{\left( 0\ \text{m/s} \right)^2+2\left( -9.81\ \text{m/s}^2 \right)\left( 0 \ \text{m}-75\ \text{m} \right)} \\ v_2 & = \pm \ 38.4\ \text{m/s}\\ v_2 & =- 38.4\ \text{m/s}\\ \end{align*}

Since the object is directing downwards when it hit the ground, the velocity is negative.

Part C

First, we calculate the total time of the object’s motion from the beginning to the ground.

\begin{align*} \Delta y & =\bcancel{v_{oy}t}+ \frac{1}{2}at^2 \\ \Delta y & = \frac{1}{2}at^2 \\ 0 \text{m}-75\ \text{m} & = \frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)t^2 \\ t^2& =\frac{-75\ \text{m}}{-4.905 \text{m/s}^2}\\ t&=\sqrt{\frac{75\ \text{m}}{4.905 \text{m/s}^2}}\\ t&=3.91 \ \text{s} \end{align*}

Second, determine the total distance traveled from 0 s to 2.91 s, leaving out the last 1 s of the motion.

\begin{align*} \Delta y & = v_{oy} t + \frac{1}{2}at^2\\ \Delta y & = \left( 0 \ \text{m/s} \right)\left( 1 \ \text{s} \right)+\frac{1}{2}\left( -9.81 \ \text{m/s}^2 \right)\left( 2.91 \ \text{s} \right)^2 \\ \Delta y & = 0 -41.5 \ \text{m} \\ \Delta y & = -41.5 \ \text{m} \\ |\Delta y| & =41.5 \ \text{m} \end{align*}

Finally, subtract this distance from the total distance traveled to get the distance traveled in the last 1 second.

\begin{align*} y_{_{\text{last 1 sec}}} & = 75.0 \ \text{m}-41.5 \ \text{m} \\ y_{_{\text{last 1 sec}}} & = 33.5\ \text{m} \end{align*}

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Acceleration of a Revolving Ball – Uniform Circular Motion Example

A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m, as in the Figure 1 below. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?

Figure 1: A small object moving in a circle, showing how the velocity changes. At each point, the instantaneous velocity is in a direction tangent to the circular path.

Solution:

The linear velocity of the ball can be computed by dividing the total arc length traveled by the total time of travel. That is, the ball traveled 2 revolutions (twice the circumference of the circle) for 1 second. Thus,

\begin{align*} \text{v} &= \frac{2\cdot2 \pi \text{r}}{\text{t}} \\ \\ & = \frac{4\pi \text{r}}{\text{t}} \\ \\ & = \frac{4\pi\left( 0.600\ \text{m} \right)}{1 \ \text{s}} \\ \\ & = 7.54 \ \text{m/s} \end{align*}

Since the linear velocity has already been computed, we can now compute for the centripetal acceleration, a_c.

\begin{align*} \text{a}_\text{c} & = \frac{\text{v}^{2}}{\text{r}} \\ \\ & = \frac{\left( 7.54\ \text{m/s} \right)^{2}}{0.600\ \text{m}}\\ \\ & =94.8 \ \text{m/s}^{2} \end{align*}

College Physics 2.51 – Time of the hiker to move out from a falling rock

Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?

Solution:

Part A

We know that the initial height, $y_0$ of the rock is 105 meters, and the initial velocity, $v_0$ is zero. We shall solve for the distance traveled by the rock for 1.5 seconds from the initial position first to find the height at detection.

The change in height is

\displaystyle \begin{aligned} \Delta \text{y}&=\text{v}_0\text{t}+\frac{1}{2}\text{at}^2 \\ &=\left( 0 \right)\left( 1.50 \ \text{s} \right)+\frac{1}{2}\left( 9.81\ \text{m/s}^{2} \right)\left( 1.50\ \text{s} \right)^{2}\\ &=0+11.036\ \text{m} \\ &=11.04 \ \text{m} \end{aligned}

So, the rock falls about 11.04 m from the initial height for 1.50 seconds. Therefore, the height of the rock above his head at this point is

\displaystyle \begin{aligned} \text{y}&=\text{y}_{0}-\Delta \text{y} \\ &=105\ \text{m}-11.04\ \text{m} \\ &=93.96 \ \text{m} \end{aligned}

Part B

We shall solve for the total time of travel, that is, from the initial position to his head. Then we shall subtract 1.50 s from that to solve for the unknown time of moving out. The total time of travel is

\begin{aligned} \text{y} & =\frac{1}{2}\text{at}^{2} \\ &\text {Solving for t, we have}\\ \text{t}&=\sqrt{\frac{\text{2y}}{\text{a}}} \\ &=\sqrt{\frac{2\left( 105\ \text{m} \right)}{9.81 \ \text{m/s}^{2}}} \\ &=4.63 \ \text{s} \end{aligned}

Therefore, to move out the hiker has about

\begin{aligned} \text{t}&=4.63 \ \text{s}-1.50\ \text{s}\\ &=3.13\ \text{s} \end{aligned}

College Physics 2.50 – Motion of a Jumping Kangaroo

A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?

Part A

The motion of the kangaroo is under free-fall. We are looking for the initial velocity, and we know that the velocity in the highest position is zero.

From

\begin{aligned} \text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay},\\ \end{aligned}

we have

\begin{aligned} \text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay}\\ \text{v}^2-2\text{ay} &= \left ( \text{v}_0\right)^2\\ \text{v}_0&=\sqrt{\text{v}^2-2\text{ay}} \end{aligned}

Substituting the known values,

\begin{aligned} \text{v}_0&=\sqrt{\text{v}^2-2\text{ay}} \\ \text{v}_0&=\sqrt{0^2-2\left(-9.81 \text{m/s}^2\right)\left(2.50 \text{m}\right)}\\ \text{v}_0&= {\color{green}7.00 \ \text{m/s}} \end{aligned}

Therefore, the vertical speed of the kangaroo when it leaves the ground is 7.00 m/s.

Part B

Since the motion of the kangaroo has uniform acceleration, we can use the formula

\text{y}=\text{v}_o\text{t}+\frac{1}{2}\text{a}\text{t}^2

The initial and final position of the kangaroo will be the same, so $y$ is equal to zero. The initial velocity is 7.00 m/s, and the acceleration is -9.81 m/s².

\begin{aligned} \text{y} & =\text{v}_0\text{t}+\frac{1}{2}\text{a}\text{t}^2\\ 0 & = \left( 7.00\ \text{m/s} \right)\text{t}+\frac{1}{2}\left( -9.81\ \text{m/s}^{2} \right)\text{t}^2\\ 0 & =7\text{t}-4.905\text{t}^{2}\\ 7\text{t}-4.905\text{t}^{2}&=0 \\ \text{t}\left( 7-4.905\text{t} \right) & =0 \\ \text{t}=0 \qquad &\text{or} \qquad 7-4.905\text{t}=0 \\ \end{aligned}

Discard the time 0 since this refers to the beginning of motion. Therefore, we have

\begin{aligned} 7-4.905\text{t} &=0 \\ 4.905\text{t} & = 7 \\ \text{t} & =\frac{7}{4.905} \\ \text{t}&={\color{green}1.43 \ \text{s}} \end{aligned}

The kangaroo is about 1.43 seconds long in the air.

$Solution Guides to College Physics by Openstax Chapter 13 Banner$

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A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?

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Thermal Expansion of Solids and Liquids

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