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College Physics by Openstax Chapter 2 Problem 53

There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff.
(a) How fast will it be going when it strikes the ground?
(b) Assuming a reaction time of 0.300 s, how long will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is 335 m/s on this day.

Solution:

Part A

Consider the Figure A. We shall consider two position points — position 1 at the top of the cliff, and position 2 on the ground.

Position 1 is 250 meters above the ground (y=250 m), and since this is the initial position, t=0, and the initial velocity is vy=0.

Position 2 is on the ground (y=0), and we do not know the time and velocity at this point.

For this part, we will solve for the value of vy at position 2. 

\begin{align*}
\left( v_{y_{2}} \right)^2 &  = \left( v_{y_{1}} \right)^2+2a \Delta y \\
\left( v_{y_{2}} \right)^2 &  = \left( 0\ \text{m/s} \right)^2+2\left( -9.81\ \text{m/s}^2 \right)\left( 0\ \text{m}-250\ \text{m} \right) \\
\left( v_{y_{2}} \right)^2 &  = 4905\ \text{m}^2/\text{s}^2 \\
v_{y_{2}} & = \pm \sqrt{4905\ \text{m}^2/\text{s}^2} \\
v_{y_{2}} & = \pm \ 70.0\ \text{m/s}\\
v_{y_{2}} & = -  70.0\ \text{m/s} \qquad {\color{gold}\left( \text{Answer} \right) }\\
\end{align*}
Figure A

Since the boulder is moving downward at this point, we shall consider the negative sign of the velocity as an indication of the downward direction of the velocity.

To answer the question, the boulder is going at about 70.0 m/s downward when it strikes the ground.

Part B

Let:
t_s be the time for the sound to travel from the top of the cliff to the tourist;
t be the total time elapsed before the tourist can react;
t_b be the time for the rock to travel from the top of the cliff to the ground
t_t be the amount of time the tourist has to get out of the way after hearing the sound of the rock breaking loose

Since we are given the speed of sound, we have

\begin{align*}

t_s & = \frac{\text{height of the cliff}}{\text{speed of sound}} \\
t_s & = \frac{250\ \text{m}}{335\ \text{m/s}}\\
t_s & =0.746 \ \text{s}

\end{align*}

So, the tourist can react after

\begin{align*}

t & =t_s + \text{reaction time} \\
t & = 0.746\ \text{s}+0.300\ \text{s} \\
t &= 1.046\ \text{s}

\end{align*}

Then, we can compute for the total time it takes for the rock to travel from top to the ground. 

\begin{align*}

t_b & =\frac{v_{y_2}-v_{y_1}}{a} \\
t_b & =\frac{-70\ \text{m/s}-0\ \text{m/s}}{-9.81\ \text{m/s}^2} \\
t_b & = 7.14\ \text{s}

\end{align*}

Therefore, the tourist still has the time to get out of the way. That is,

\begin{align*}

t_t & =7.14\ \text{s}-1.046\ \text{s} \\
t_t & = 6.09\ \text{s} \qquad {\color{gold}\left( \text{Answer} \right) } \\

\end{align*}

\therefore the tourist has about 6.09 seconds to get out of the way.

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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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Skidding on a Curve: Unbanked Curves – Uniform Circular Motion Example Problem

A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s Will the car follow the curve, or will it skid? Assume: (a) the pavement is dry and the coefficient of static friction is \mu _s=0.60; (b) the pavement is icy and \mu _s=0.25.

Solution:

The forces on the car are gravity mg downward, the normal force FN exerted upward by the road, and a horizontal friction force due to the road. They are shown in the free-body diagram of the car below. The car will follow the curve if the maximum static friction force is greater than the mass times the centripetal acceleration.

Part A

In the vertical direction (y) there is no acceleration. Newton’s second law tells us that the normal force on the car is equal to the weight mg since the road is flat:

\begin{align*}
\sum_{}^{}F_y & =m\ a_c\\
\\
F_N\ -\ mg & =0\\
\\
F_N&=mg\\
\\
F_N &=\left( 1000\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)\\
\\
F_N &=9810\  \text{N}
\end{align*}

In the horizontal direction the only force is friction, and we must compare it to the force needed to produce the centripetal acceleration to see if it is sufficient. The net horizontal force required to keep the car moving in a circle around the curve is

\begin{align*}
\sum_{}^{}F_c & =m\ a_c\\
\\
 & =m\cdot \frac{v^2}{r}\\
\\
& =\left( 1000\ \text{kg} \right)\cdot \frac{\left( 15\ \text{m/s} \right)^2}{50\ \text{m}}\\
\\
&=4500\ \text{N}
\end{align*}

Now we compute the maximum total static friction force (the sum of the friction forces acting on each of the four tires) to see if it can be large enough to provide a safe centripetal acceleration. For (a), \mu _s=0.60, and the maximum friction force attainable is

\begin{align*}
\sum_{}^{}F_{fr_{max}}& =\mu _s \ F_N\\
\\
&=\left( 0.60 \right)\left( 9810\ \text{N} \right)\\
\\
&=5886\ \text{N}
\end{align*}

Since a force of only 4500 N is needed, and that is, in fact, how much will be exerted by the road as a static friction force, the car can follow the curve.

Part B

The maximum static friction force possible is

\begin{align*}
\sum_{}^{}F_{fr_{max}}& =\mu _s \ F_N\\
\\
&=\left( 0.25 \right)\left( 9810\ \text{N} \right)\\
\\
&=2452.5\ \text{N}
\end{align*}

The car will skid because the ground cannot exert sufficient force (4500 N is needed) to keep it moving in a curve of radius 50 m at a speed of 54 km/h.

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

Chapter 13: Temperature, Kinetic Theory, and the Gas Laws

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Thermal Expansion of Solids and Liquids

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The Ideal Gas Law

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Humidity, Evaporation, and Boiling

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Chapter 12: Fluid Dynamics and Its Biological and Medical Applications

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Bernoulli’s Equation

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The Onset of Turbulence

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Chapter 11: Fluid Statics

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Chapter 10: Rotational Motion and Angular Momentum

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Chapter 9: Statics and Torque

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The Second Condition for Equilibrium

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Stability

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Simple Machines

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Forces and Torques in Muscles and Joints

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