College Physics 3.30 – A rugby player passes the ball across the field


A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand.

(a) At what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used?

(b) What other angle gives the same range, and why would it not be used?

(c) How long did this pass take?


Solution:

To illustrate the problem, consider the following figure:

A player passes the ball 7 meters across the field with an initial velocity of 12 m/s and directed at an angle θ from the horizontal.
A player passes the ball 7 meters across the field with an initial velocity of 12 m/s

Part A

We are given the 7-meter range, R, and the initial velocity, vo, of the projectile. We have R=7.0 m, and vo=12.0 m/s. To solve for the angle of the initial velocity, we will use the formula for range

\displaystyle \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

\displaystyle \text{gR}=\text{v}_{\text{o}}^2\sin 2\theta _{\text{o}}

\displaystyle \sin 2\theta _{\text{o}}=\frac{\text{gR}}{\text{v}_{\text{o}}^2}

\displaystyle 2\theta _\text{o}=\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right)

\displaystyle \theta _\text{o}=\frac{1}{2}\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right)

Substituting the given values, we have

\displaystyle \theta _\text{o}=\frac{1}{2} \sin ^{-1}\left[\frac{\left(9.81\text{m/s}^2\right)\left(7.0\text{m}\right)}{\left(12.0\text{m/s}\right)^2}\right]

\displaystyle \theta _\text{o}=14.24^{\circ}     ◀

Part B

The other angle that would give the same range is actually the complement of the solved angle in Part A. The other angle,

\displaystyle \theta _o'=90^{\circ} -14.24^{\circ} =75.76^{\circ}     ◀

This angle is not used as often, because the time of flight will be longer. In rugby that means the defense would have a greater time to get into position to knock down or intercept the pass that has the larger angle of release.

Part C

We can use the x-component of the motion to solve for the time of flight.

\displaystyle \Delta \text{x}=\text{v}_\text{x}\text{t}

We need the horizontal component of the velocity. We should be able to solve for the component since we are already given the initial velocity and the angle.

\displaystyle \text{v}_{\text{x}}=\left(12\:\text{m/s}\right)\cos 14.24^{\circ} =11.63\:\text{m/s}

Therefore, the total time of flight is

\displaystyle \text{t}=\frac{\Delta \text{x}}{\text{v}_{\text{x}}}

\displaystyle \text{t}=\frac{7.0\:\text{m}}{11.63\:\text{m/s}}

\displaystyle \text{t}=0.60\:\text{s}     ◀


College Physics 3.29 – An archer shoots an arrow at a distant target


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
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

\displaystyle \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

\displaystyle \text{gR}=\text{v}_{\text{o}}^2\:\sin 2\theta _{\text{o}}

\displaystyle \sin \:2\theta _{\text{o}}=\frac{\text{gR}}{\text{v}_{\text{o}}^2}

\displaystyle 2\theta _\text{o}=\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right)

\displaystyle \theta _\text{o}=\frac{1}{2}\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right)

Substituting the given values, we have

\displaystyle \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]

\displaystyle \theta _o=18.46^{\circ}     ◀

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

\displaystyle \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

\displaystyle \text{v}_{\text{oy}}=\left(35.0\:\text{m/s}\right)\sin 18.46^{\circ}

\displaystyle \text{v}_{\text{oy}}=11.08\:\text{m/s}

Therefore, the maximum height is

\displaystyle \text{h}_{\max }=\frac{\left(11.08\:\text{m/s}\right)^2}{2\left(9.81\:\text{m/s}^2\right)}

\displaystyle \text{h}_{\max }=6.26\:\text{m}     ◀

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


College Physics 3.28 – A daredevil attempting to jump his motorcycle


(a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32º ramp at a speed of 40.0 m/s (144 km/h) . How many buses can he clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 20.0 m long?

(b) Discuss what your answer implies about the margin of error in this act—that is, consider how much greater the range is than the horizontal distance he must travel to miss the end of the last bus. (Neglect air resistance.)


Solution:

To illustrate the problem, consider the following figure:

The projectile path of the daredevil from the ramp. The ramp is inclined at 32° from the horizontal.
The projectile path of the daredevil from the ramp

Part A

To determine the number of buses that the daredevil can clear, we will divide the range of the projectile path by 20 m, the length of 1 bus. That is

\displaystyle \text{no. of bus}=\frac{\text{Range}}{\text{bus length}}

First, we need to solve for the range.

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

\displaystyle \text{Range}=\frac{\left(40.0\:\text{m/s}\right)^2\sin \left[2\left(32^{\circ} \right)\right]}{9.81\:\text{m/s}^2}

\displaystyle \text{Range}=146.7\:\text{m}

Therefore, the number of buses cleared is

\displaystyle \text{no. of buses}=\frac{146.7\:\text{m}}{20\:\text{m}}

\displaystyle \text{no. of buses}=7.34\:\text{buses}

Therefore, he can only clear 7 buses.          ◀

Part B

He clears the last bus by 6.7 m, which seems to be a large margin of error, but since we neglected air resistance, it really isn’t that much room for error.


College Physics 3.27 – A ball thrown from the top of a building


A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance.

(a) How long is the ball in the air?

(b) What must have been the initial horizontal component of the velocity?

(c) What is the vertical component of the velocity just before the ball hits the ground?

(d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?


Solution:

To illustrate the problem, consider the following figure:

The path of the ball thrown at the top of a 60 m building. The horizontal displacement is 100 meters.
The path of the ball thrown at the top of a 60 m building.

Part A

The problem states that the initial velocity is horizontal, this means that the initial vertical velocity is zero. We are also given the height of the building (which is a downward displacement), so we can solve for the time of flight using the formula y=voyt+1/2at2. That is,

\displaystyle \text{y}=\text{v}_{\text{oy}}\text{t}+\frac{1}{2}\text{a}\text{t}^2

\displaystyle -60\:\text{m}=0+\frac{1}{2}\left(-9.81\:\text{m/s}^2\right)\text{t}^2

\displaystyle \:\text{t}^2=\frac{-60\:\text{m}}{-4.905\:\text{m/s}^2}

\displaystyle \:\text{t}^2=12.2324\:\text{s}^2

\displaystyle \text{t}=3.50\:\text{s}          ◀

Part B

To solve for the vox, we will use the formula \displaystyle \text{v}_{\text{ox}}=\frac{\Delta \:\text{x}}{\text{t}}.

\displaystyle \text{v}_{\text{ox}}=\frac{100\:\text{m}}{3.50\:\text{s}}

\displaystyle \text{v}_{\text{ox}}=28.57\:\text{m/s}          ◀

Part C

To solve for the velocity as the ball hits the ground, we shall consider two points: (1) at the beginning of the flight, and (2) when the ball hits the ground.

We know that the initial velocity, voy, is zero. To solve for the final velocity, we will use the formula \displaystyle \text{v}_{\text{f}}=\text{v}_{\text{o}}+\text{at}

\displaystyle \text{v}_{\text{f}}=0+\left(-9.81\:\text{m/s}^2\right)\left(3.50\:\text{s}\right)

\displaystyle \text{v}_{\text{f}}=-34.34\:\text{m/s}          ◀

The negative velocity indicates that the motion is downward.

Part D

Since we already know the horizontal and vertical components of the velocity when it hits the ground, we can find the resultant.

\displaystyle \text{v}=\sqrt{\text{v}_{\text{x}}^2+\text{v}_{\text{y}}^2}

\displaystyle \text{v}=\sqrt{\left(28.57\:\text{m/s}\right)^2+\left(-34.34\:\text{m/s}\right)^2}

\displaystyle \text{v}=44.67\:\text{m/s}          ◀

The direction of the velocity is

\displaystyle \theta {\text{x}}=\tan ^{-1}\left|\frac{\text{v}_{\text{y}}}{\text{v}_{\text{x}}}\right|

\displaystyle \theta _{\text{x}}=\tan ^{-1}\left|\frac{-34.34}{28.57}\right|

\displaystyle \theta _{\text{x}}=50.24^{\circ} ⦪          ◀

The velocity is directed 50.24° down the x-axis.


College Physics 3.26 – A ball kicked with x and y initial velocities


A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction.

(a) At what speed does the ball hit the ground?

(b) For how long does the ball remain in the air?

(c)What maximum height is attained by the ball?


Solution:

To illustrate the problem, consider the following figure:

The path of the projectile with initial horizontal velocity of 16 m/s and vertical velocity of 12 m/s.
The path of the projectile with initial horizontal and vertical velocities given.

Part A

Since the starting position has the same elevation as when it hits the ground, the speeds at these points are the same. The final speed is computed by solving the resultant of the horizontal and vertical velocities. That is

\displaystyle \text{v}_{\text{f}}=\sqrt{\left(\text{v}_{\text{ox}}\right)^2+\left(\text{v}_{\text{oy}}\right)^2}

\displaystyle \text{v}_{\text{f}}=\sqrt{\left(16\:\text{m/s}\right)^2+\left(12\:\text{m/s}\right)^2}

\displaystyle \text{v}_{\text{f}}=\sqrt{400\:\text{(m/s)}^2}

\displaystyle \text{v}_{\text{f}}=20\:\text{m/s}          ◀

Part B

Consider the two points: (1) the starting point and (2) the highest point.

We know that at the highest point, the vertical velocity is zero. We also know that the total time of the flight is twice the time from the beginning to the top.

So, we shall use the formula \displaystyle \text{t}=\frac{\text{v}_{\text{f}}-\text{v}_{\text{o}}}{\text{a}}.

\displaystyle \text{t}=2\left(\frac{\text{v}_{\text{top}}-\text{v}_{\text{o}}}{\text{a}}\right)

\displaystyle \text{t}=2\left(\frac{0\:\text{m/s}-12\:\text{m/s}}{-9.81\:\text{m/s}^2}\right)

\displaystyle \text{t}=2.45\:\text{s}          ◀

Part C

The maximum height attained can be calculated using the formula \displaystyle \left(\text{v}_{\text{f}}\right)^2=\left(\text{v}_{\text{o}}\right)^2+2\text{a}\text{y}.

The maximum height is calculated as follows:

\displaystyle \left(\text{v}_{\text{f}}\right)^2=\left(\text{v}_{\text{o}}\right)^2+2\text{ay}

\displaystyle \text{y}_{\max }=\frac{\left(\text{v}_{\text{top}}\right)^2-\left(\text{v}_{\text{o}}\right)^2}{2\text{a}}

\displaystyle \text{y}_{\max }=\frac{\left(0\:\text{m/s}\right)^2-\left(16\:\text{m/s}\right)^2}{2\left(-9.81\:\text{m/s}^2\right)}

\displaystyle \text{y}_{\max }=7.34\:\text{m}          ◀


College Physics 3.25 – A projectile launched at ground level


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. The projectile has an initial velocity of 50 m/s and directed 30° from the horizontal.
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.

\displaystyle \text{v}_{\text{ox}}=\left(50\:\text{m/s}\right)\cos 30^{\circ} =43.3013\:\text{m/s}

\displaystyle \text{v}_{\text{oy}}=\left(50\:\text{m/s}\right)\sin 30^{\circ} =25\:\text{m/s}

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,

\displaystyle \text{x}=\text{v}_{\text{ox}}\text{t}

\displaystyle \text{x}=\left(43.3013\:\text{m/s}\right)\left(3\:\text{s}\right)

\displaystyle \text{x}=129.9\:\text{m}          ◀        

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

\displaystyle \text{y}=\text{v}_{\text{oy}}\text{t}+\frac{1}{2}\text{a}\text{t}^2

\displaystyle \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

\displaystyle \text{y}=30.9\:\text{m}          ◀

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


Farmer wants to Fence off his Four-Sided Plot with missing Side| Vector Addition and Subtraction| Analytical Method| College Physics| Problem 3.22


A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in the figure, and then correctly calculates the length and orientation of the fourth side D. What is his result?

8

SOLUTION:

We know that the sum of the 4 vectors is zero. So, Vector D is calculated as

D=-A-B-C

We solve for the x-component first. The x-component of vector D is

D_x=-A_x-B_x-C_x

D_x=-\left(4.70\:km\right)cos\left(-7.50^{\circ} \right)-\left(2.48\:km\right)cos106^{\circ} -\left(3.02\:km\right)cos161^{\circ}

D_x=-1.12\:km

We solve for the y-component.

D_y=-A_y-B_y-C_y

D_y=-\left(4.70\:km\right)sin\left(-7.50^{\circ} \right)-\left(2.48\:km\right)sin106^{\circ} -\left(3.02\:km\right)sin161^{\circ}

D_y=-2.75\:km

Since we already know the x and y component of vector D, we can finally solve for the distance of vector D.

D=\sqrt{\left(D_x\right)^2+\left(D_y\right)^2}

D=\sqrt{\left(-1.12\:km\right)^2+\left(-2.75\:km\right)^2}

D=2.97\:km

The corresponding direction of vector D is

\theta =tan^{-1}\left|\frac{D_x}{D_y}\right|

\theta =tan^{-1}\left|\frac{1.12\:km}{2.75\:km}\right|

\theta =22.2^{\circ} \:W\:of\:S


College Physics 3.21 – Vector Addition and Subtraction


You fly 32.0 km in a straight line in still air in the direction 35.0° south of west.

(a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.)

(b) Find the distances you would have to fly first in a direction 45.0º south of west and then in a direction 45.0° west of north. These are the components of the displacement along a different set of axes—one rotated 45.0°. 


Solution:

Part A

The component along the south direction is

D_s=Rsin\theta =\left(32.0\:km\right)sin\:35^{\circ} =18.4\:km

The component along the west direction is

D_w=Rcos\theta =\left(32.0\:km\right)cos35^{\circ} =26.2\:km

Part B

Consider the following figure with the rotated axes x’-y’.

7

The component along the southwest direction is

D_{SW}=Rcos\theta '=\left(32.0\:km\right)cos10^{\circ} =31.5\:km

The component along the northwest direction is

D_{NW}=Rsin\theta '=\left(32.0\:km\right)cos10^{\circ} =5.56\:km


College Physics 3.20 – Triangular Piece of Flat Land


A new landowner has a triangular piece of flat land she wishes to fence. Starting at the west corner, she measures the first side to be 80.0 m long and the next to be 105 m. These sides are represented as displacement vectors A from B in Figure 3.61. She then correctly calculates the length and orientation of the third side C. What is her result?

The displacement vectors A, B, and C formed a closed triangle. A is 80 m directed 21° clockwise from the positive x-axis, vector B has a magnitude of 105 m and is directed 11° from the positive y-axis. Find for the magnitude and direction of vector C.
Figure 3.61: The displacement vectors A, B, and C.

Solution:

Since the figure is closed, we know that the sum of the three vectors is equal to 0. So, we have 

\vec{A}+\vec{B}+\vec{C}=0

Solving for C, we have

\vec{C}=-\vec{A}-\vec{B}

Next, we solve for the x and y components of Vectors A and B. 

For Vector A, the x and y components are

A_x=\left(80\:m\right)cos\left(21^{\circ} \right)=74.6864\:m

A_y=-\left(80\:m\right)sin\left(21^{\circ} \right)=-28.6694\:m

For Vector B, the x and y components are

B_x=-\left(105\:m\right)sin\:11^{\circ} =-20.0349\:m

B_y=\left(105\:m\right)cos\:11^{\circ} =103.0709\:m

Solve for the components of Vector C. 

C_x=-A_x-B_x=-74.6864\:m\:-\left(-20.0349\:m\right)=-54.6515\:m

C_y=-A_y-B_y=-\left(-28.6694\:m\right)-\left(103.0709\:m\right)=-74.4015\:m

Therefore, the length of Vector C is

C=\sqrt{\left(C_x\right)^2+\left(C_y\right)^2}

C=\sqrt{\left(-54.6515\:m\right)^2+\left(-74.4015\:m\right)^2}

C=92.3167\:m

The direction of Vector C is

\theta =tan^{-1}\left(\frac{C_y}{C_x}\right)=tan^{-1}\left(\frac{74.4015\:m}{54.6515\:m}\right)=53.7^{\circ} \:\:South\:of\:West


College Physics 3.18 – Driving in a straight line vs east and north distances


You drive 7.50 km in a straight line in a direction 15° east of north.

(a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.)

(b) Show that you still arrive at the same point if the east and north legs are reversed in order.


Solution:

Part A

Consider the following figure:

North and East Components of the given displacement
North and East Components of the given displacement

The east distance is the component in the horizontal direction.

D_E=7.50\:km\:\cdot sin\:\left(15^{\circ} \right)

D_E=1.94\:km

The north distance is the vertical component

D_E=7.50\:km\cdot cos\left(15^{\circ} \right)

D_E=7.24\:km

Part B

3.19

Based from the figure, we can easily see that the order is reversible in the addition of vectors. We say that D_E+D_N=D_N+D_E