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College Physics by Openstax Chapter 3 Problem 35

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Solution:

We are required to solve for the distance in a standing broad jump. To do this, we can use the formula for the range of a projectile motion. However, we need the following assumptions:

  • The jumper leaves the ground in a 45° angle from the horizontal, for maximum horizontal displacement.
  • The jumper is on level ground.

The formula for the range is 

\text{R}=\frac{\text{v}_{\text{o}}^2\:\sin 2\theta _{\text{o}}}{\text{g}}

To find the initial velocity of the jump, vo, we shall use the kinematic formula from the crouch position to the time the person leaves the ground. 

\text{v}_{\text{f}}^2=\text{v}_{\text{o}}^2+2\text{ax}

In this case, the final velocity will be the initial velocity of the jump. 

\begin{align*}
 \text{v}_{\text{f}}=\sqrt{\left(0\:\text{m/s}\right)^2+2\left(1.25\times 9.81\:\text{m/s}^2\right)\left(0.600\:\text{m}\right)}=3.84\:\text{m/s}
\end{align*}

So, the initial velocity of the flight is 3.84 m/s. We can now use the formula for range.

\begin{align*}
\text{R}&=\frac{\text{v}_{\text{o}}^2\:\sin 2\theta_{\text{o}}}{\text{g}} \\
\text{R}&=\frac{\left(3.84\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}\\
\text{R}&=1.50\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 3 Problem 34

An arrow is shot from a height of 1.5 m toward a cliff of height H . It is shot with a velocity of 30 m/s at an angle of 60º above the horizontal. It lands on the top edge of the cliff 4.0 s later.
(a) What is the height of the cliff?
(b) What is the maximum height reached by the arrow along its trajectory?
(c) What is the arrow’s impact speed just before hitting the cliff?

Solution:

Consider the following illustration:

An arrow shot at a height of 1.5 m towards a cliff of height H

Part A

We are required to solve for the value of H. We shall use the formula

\Delta \text{y}=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2

or, we can also write the formula as

 \text{y}-\text{y}_0=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2

Substituting the given values, we have

\begin{align*}
 \text{y}-\text{y}_0 &=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2 \\

\text{H}-1.5\:\text{m} & =\left(30\:\text{m/s}\:\sin 60^{\circ} \right)\left(4.0\:\text{s}\right)+\frac{1}{2}\left(-9.81\:\text{m/s}^2\right)\left(4.0\:\text{s}\right)^2 \\

\text{H}-1.5\:\text{m} & =25.44\:\text{m} \\

\text{H} & =25.44\:\text{m}+1.5\:\text{m} \\

\text{H} & =26.94\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Part B

The maximum height of the projectile is given by the formula

\Delta \text{y}=\frac{\text{v}_{\text{0y}}^2}{2\text{g}}

or the formula can be written as

\text{y}_{\text{max}}-\text{y}_{\text{0}}=\frac{\text{v}_{\text{0y}}^2}{-2\text{g}}

Therefore, we have

\begin{align*}
\text{y}_{\text{max}}-\text{y}_{\text{0}} & =\frac{\text{v}_{\text{0y}}^2}{-2\text{g}} \\

\text{y}_{\text{max}}& =\frac{\text{v}_{\text{0y}}^2}{-2\text{g}}+\text{y}_{\text{0}} \\

\text{y}_{\text{max}}&=\frac{\left(\left(30\:\text{m/s}\right)\sin 60^{\circ} \right)^2}{-2\left(-9.81\:\text{m/s}^2\right)}+1.5\:\text{m} \\

\text{y}_{\text{max}}&=34.40\:\text{m}+1.5\:\text{m} \\

\text{y}_{\text{max}} & =35.90\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}
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Part C

To solve for the final speed, we need the vertical and horizontal components when the arrow hits the cliff.

We are going to use the formula for acceleration along the vertical to solve for the final speed in the vertical direction. That is

\text{a}=\frac{\text{v}_{\text{y}}-\text{v}_{\text{0y}}}{\text{t}}

Solving for vfy in terms of the other variables, we have

\begin{align*}
\text{v}_{\text{y}}&=\text{v}_{\text{0y}}+\text{at}\\
\text{v}_{\text{y}}&=\left(30\:\text{m/s}\right)\sin 60^{\circ} +\left(-9.81\:\text{m/s}^2\right)\left(4.0\:\text{s}\right) \\
 \text{v}_{\text{y}}&=-13.25\:\text{m/s}
\end{align*}

Since we know that the horizontal component of the velocity does not change along the entire flight, we can equate the initial and final horizontal velocities. That is 

\begin{align*}
\text{v}_{\text{x}}&=\text{v}_{\text{0x}}=\left(30\:\text{m/s}\right)\cos 60^{\circ} \\
\text{v}_{\text{x}}&=15\:\text{m/s}
\end{align*}

Therefore, the final speed is

\begin{align*}
\text{v}&=\sqrt{\text{v}_{\text{y}}^2+\text{v}_{\text{x}}^2} \\
\text{v}&=\sqrt{\left(-13.25\:\text{m/s}\right)^2+\left(15\:\text{m/s}\right)^2}\\
\text{v}&=20.01\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 3 Problem 32

Verify the ranges shown for the projectiles in Figure 3.40(b) for an initial velocity of 50 m/s at the given initial angles.

Solution:

To verify the given values in the figure, we need to solve for individual ranges for the given initial angles. To do this, we shall use the formula

\displaystyle \text{R}=\frac{\text{v}_{\text{0}}^2 \sin 2\theta _{\text{0}}}{\text{g}}

When the initial angle is 15°, the range is

\displaystyle \text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 15^{\circ} \right)}{9.81\:\text{m/s}^2}=127.42\:\text{m}

When the initial angle is 45°, the range is

\displaystyle \text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=254.84\:\text{m}

When the initial angle is 75°, the range is

\displaystyle \text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 75^{\circ} \right)}{9.81\:\text{m/s}^2}=127.42\:\text{m}

Based on the result of the calculations, we can say that the numbers in the figure are verified. The very small differences are only due to round-off errors.

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College Physics by Openstax Chapter 3 Problem 29

An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

Solution:

To illustrate the problem, consider the following figure:

The archer and the target at 75 meter range

Part A

We are given the range of 75-meter range, R, and the initial velocity, vo, of the projectile. We have R=75.0 m, and vo=35.0 m/s. To solve for the angle of the initial velocity, we will use the formula for range

\text{R}=\frac{\text{v}^{2}_{\text{o}}\:\sin 2\theta _{\text{o}}}{g}

Solving for θo in terms of the other variables, we have

\begin{align*}

\text{gR} & =\text{v}_{\text{o}}^2\:\sin 2\theta _{\text{o}} \\
\sin \:2\theta _{\text{o}} & =\frac{\text{gR}}{\text{v}_{\text{o}}^2} \\
2\theta _\text{o} & =\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right) \\
\theta _\text{o} & =\frac{1}{2}\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right) \\
\theta _o & =\frac{1}{2}\sin ^{-1}\left[\frac{\left(9.81\:\text{m/s}^2\right)\left(75.0\:\text{m}\right)}{\left(35.0\:\text{m/s}\right)^2}\right] \\
\theta _o & =18.46^{\circ} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
 
\end{align*}

Part B

We know that halfway, the maximum height of the projectile occurs. Also at this instant, the vertical velocity is zero. We can solve for the maximum height and compare it with the given height of 3.50 meters.

The maximum height can be computed using the formula

\text{h}_{\text{max}}=\frac{\text{v}_{\text{oy}}^2}{2\text{g}}

To compute for the maximum height, we need the initial vertical velocity, voy. Since we know the magnitude and direction of the initial velocity, we have

\begin{align*}

\text{v}_{\text{oy}} & =\left(35.0\:\text{m/s}\right)\sin 18.46^{\circ} \\
\text{v}_{\text{oy}} & =11.08\:\text{m/s}
 
\end{align*}

Therefore, the maximum height is

\begin{align*}

\text{h}_{\max } & =\frac{\left(11.08\:\text{m/s}\right)^2}{2\left(9.81\:\text{m/s}^2\right)} \\
\text{h}_{\max } & =6.26\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}

 
\end{align*}

We have known that the path of the arrow is above the branch of the tree. Therefore, the arrow will go through.

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College Physics by Openstax Chapter 3 Problem 25

A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0º above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the x and y distances from where the projectile was launched to where it lands?

Solution:

Since we do not know the exact location of the projectile after 3 seconds, consider the following arbitrary figure:

The path of the projectile from the ground to a point 3 seconds later.

From the figure, we can solve for the components of the initial velocity.

\begin{align*}
\text{v}_{\text{ox}} &=\left(50\:\text{m/s}\right)\cos 30^{\circ} \\
& =43.3013\:\text{m/s}
\\
\\
\text{v}_{\text{oy}} & =\left(50\:\text{m/s}\right)\sin 30^{\circ} \\
&=25\:\text{m/s}
\\
\end{align*}

So, we are asked to solve for the values of x and y. To solve for the value of the horizontal displacement, x, we shall use the formula x=voxt. That is,

\begin{align*}
\text{x} & =\text{v}_{\text{ox}}\text{t} \\
\text{x} & =\left(43.3013\:\text{m/s}\right)\left(3\:\text{s}\right) \\
\text{x} & =130\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

To solve for the vertical displacement, y, we shall use the formula y=voyt+1/2at2. That is

\begin{align*}
\text{y} & =\text{v}_{\text{oy}}\text{t}+\frac{1}{2}\text{a}\text{t}^2 \\
\text{y} & =\left(25\:\text{m/s}\right)\left(3\:\text{s}\right)+\frac{1}{2}\left(-9.81\:\text{m/s}^2\right)\left(3\:\text{s}\right)^2 \\
\text{y} & =30.9\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Therefore, the projectile strikes a target at a distance 129.9 meters horizontally and 30.9 meters vertically from the launching point.

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College Physics by Openstax Chapter 3 Problem 24

Suppose a pilot flies 40.0 km in a direction 60º north of east and then flies 30.0 km in a direction 15º north of east as shown in Figure 3.61. Find her total distance R from the starting point and the direction θ of the straight-line path to the final position. Discuss qualitatively how this flight would be altered by a wind from the north and how the effect of the wind would depend on both wind speed and the speed of the plane relative to the air mass.

Figure 3.61

Solution:

The pilot’s displacement is characterized by 2 vectors, A and B, as depicted in Figure 3.61. To determine her total displacement R from the starting point, we need to add the two given vectors. To do this, we individually get the x and y components of each vector. This is presented in the table that follows:

Vectorx-componenty-component
A40\:\cos 60^{\circ} =20\:\text{km} 40\:\sin 60^{\circ} =34.6410\:\text{km}
B 30\:\cos 15^{\circ} =28.9778\:\text{km} 30\:\sin 15^{\circ} =7.7646\:\text{km}
Sum 48.9778\: \text{km} 42.4056 \:\text{km}

The table above indicates east and north as positive components, while west and south indicate negative components. The last row is the sum of the components. These are also the x and y components of the resultant vector.

To calculate the magnitude of the resultant, we simply use the Pythagorean Theorem as follows:

\begin{align*}
\text{R} & = \sqrt{\left(48.9778\:\text{km}\right)^2+\left(42.4056\:\text{km}\right)^2} \\
\text{R} & = 64.8\:\text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\
\end{align*}

The direction of the resultant is calculated as follows:

\begin{align*}
\theta & =\tan ^{-1}\left(\frac{42.4056}{48.9778}\right) \\
\theta & =40.9^{\circ} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Therefore, the pilot’s resultant displacement is about 64.8 km directed 40.9° North of East from the starting island.

Discussion:

If the wind speed is less than the speed of the plane, it is possible to travel to the northeast, but she will travel more to the east than without the wind. If the wind speed is greater than the speed of the plane, then it is no longer possible for the plane to travel to the northeast, it will end up traveling southeast.

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College Physics by Openstax Chapter 3 Problem 12

Find the components of vtot along a set of perpendicular axes rotated 30º counterclockwise relative to those in Figure 3.55.

The figure shows v_A directed 22.5° from the positive x-axis, and v_B started from the head of v_A and is directed 23.0° from the resultant. The resultant is given to be 6.72 m/s and is directed 26.5° from v_A. In total, the resultant is measured 49° from the positive x-axis.
Figure 3.55

Solution:

By isolating the vtot from the rest of the other vectors, we come up with the following figure.

The isolated resultant velocity

The resultant velocity has a magnitude of 6.72 m/s and is directed 49° from the positive x-axis. Now, we shall create another set of axes rotated at 30° counterclockwise. We call the axes x’ and y’ axes. The figure is shown below.

The resultant velocity with the rotated axes.

From the figure, we can see that the resultant velocity is 19° from the x’ axis. Therefore, the x’ and y’ components are:

\text{x'-component}=\left(6.72\:\text{m/s}\right)\cos 19^{\circ} =6.35\:\text{m/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
 \text{y'-component}=\left(6.72\:\text{m/s}\right)\sin 19^{\circ} =2.19\:\text{m/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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College Physics by Openstax Chapter 3 Problem 11

Find the components of vtot along the x- and y-axes in Figure 3.55.

The figure shows v_A directed 22.5° from the positive x-axis, and v_B started from the head of v_A and is directed 23.0° from the resultant. The resultant is given to be 6.72 m/s and is directed 26.5° from v_A. In total, the resultant is measured 49° from the positive x-axis.
Figure 3.55

Solution:

By isolating the vtot from the rest of the other vectors, we come up with the following figure. Also, the x and y-components are shown.

The resultant velocity and its x and y components

The resultant velocity has a magnitude of 6.72 m/s and is directed 49° from the positive x-axis. To solve for the x and y components, we just need to solve the legs of the right triangle formed by the three vectors. That is, 

 \text{x-component}=\left(6.72\:\text{m/s}\right)\cos 49^{\circ} =4.41\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\text{y-component}=\left(6.72\:\text{m/s}\right)\sin 49^{\circ} =5.07\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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College Physics by Openstax Chapter 3 Problem 10

Find the magnitudes of velocities vA and vB in Figure 3.55

The figure shows v_A directed 22.5° from the positive x-axis, and v_B started from the head of v_A and is directed 23.0° from the resultant. The resultant is given to be 6.72 m/s and is directed 26.5° from v_A. In total, the resultant is measured 49° from the positive x-axis.
Figure 3.55

Solution:

Basically, we are given an oblique triangle. First, we shall determine the value of the interior angle at the intersection of vA and vB. We can solve this knowing that the sum of the interior angles of a triangle is 180°.

To solve for vA and vB, we will use the sine law. 

\begin{align*}
 \frac{\text{v}_{\text{A}}}{\sin 23^{\circ} } & =\frac{6.72\:\text{m/s}}{\sin 130.5^{\circ} } \\
\text{v}_{\text{A}} & =\frac{6.72\:\text{m/s}\:\sin \:23^{\circ }\:}{\sin \:130.5^{\circ }\:} \\
\text{v}_{\text{A}} & =3.45\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}

Using the same law to solve for the value of vB, we have

\begin{align*}
\frac{\text{v}_{\text{B}}}{\sin 26.5^{\circ} } & =\frac{6.72\:\text{m/s}}{\sin 130.5^{\circ} } \\
\text{v}_{\text{B}} & =\frac{6.72\:\text{m/s}\:\sin \:26.5^{\circ }\:\:}{\sin \:130.5^{\circ }\:} \\
\text{v}_{\text{B}} & =3.94\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}
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College Physics by Openstax Chapter 3 Problem 9

Show that the sum of the vectors discussed in Example 3.2 gives the result shown in Figure 3.24.

Figure 3.24

Solution:

So, we are given the two vectors shown below.

Vectors A and B

If we use the graphical method of adding vectors, we can join the two vectors using head-tail addition and come up with the following:

Figure 3.9B: Vectors A and B added graphically

The resultant is drawn from the tail of the first vectors (the origin) to the head of the last vector. The resultant is shown in red in the figure below.

Solve for the value of the angle 𝛼 by geometry.

\alpha = 66^\circ +\left( 180^\circ-112^\circ \right) = 134^\circ

Solve for the magnitude of the resultant using cosine law.

\begin{align*}
R^2 & = A^2+B^2-2AB\cos \alpha \\
R & = \sqrt{A^2+B^2-2AB\cos \alpha} \\
R & = \sqrt{\left( 27.5 \ \text{m} \right)^2+\left( 30.0 \ \text{m} \right)^2-2\left( 27.5\ \text{m} \right)\left( 30.0\ \text{m} \right) \cos 134^\circ} \\
R & =52.9380 \ \text{m} \\
R & = 52.9 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Solve for 𝛽 using sine law. 

\begin{align*}
\frac{\sin \beta}{B} & = \frac{\sin \alpha}{R} \\
\beta & = \sin ^{-1} \left( \frac{B \sin \alpha }{R} \right) \\
\beta & = \sin ^{-1} \left( \frac{30.0\ \text{m} \sin 134^\circ}{52.9380 \ \text{m}} \right) \\
\beta & = 24.0573^\circ
\end{align*}

Finally, solve for 𝜃.

\theta = 66^\circ+24.0573^\circ = 90.1^\circ \ \qquad \ {\color{Orange} \left( \text{Answer} \right)}

The result is in conformity with that in figure 3.24 shown on the question shown above.

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