College Physics 3.33 – A cannon fired from a battleship


The cannon on a battleship can fire a shell a maximum distance of 32.0 km.

(a) Calculate the initial velocity of the shell.

(b) What maximum height does it reach? (At its highest, the shell is above 60% of the atmosphere—but air resistance is not really negligible as assumed to make this problem easier.)

(c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is 6.37×103 km . How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here?


Solution:

Part A

We are given the range of the projectile motion. The range is 32.0 km. We also know that for the projectile to reach its maximum distance, it should be fired at 45°. So from the formula of range,

\displaystyle \text{R}=\frac{\text{v}_0^2\:\sin 2\theta _0}{\text{g}}

we can say that \sin 2\theta _0=\sin \left(2\times 45^{\circ} \right)=\sin 90^{\circ} =1. So, we have

\displaystyle \text{R}=\frac{\text{v}_0^2}{\text{g}}

We can solve for v0 in terms of the other variables. That is

\displaystyle \text{v}_0=\sqrt{\text{gR}}

Substituting the given values, we have

\displaystyle \text{v}_0=\sqrt{\left(9.81\:\text{m/s}^2\right)\left(32\times 10^3\:\text{m}\right)}

\displaystyle \text{v}_0=560.29\:\text{m/s}        ◀

Part B

We are solving for the maximum height here, which happened at the mid-flight of the projectile. The vertical velocity at this point is zero. Considering all this, the formula for the maximum height is derived to be

\displaystyle \text{h}_{\text{max}}=\frac{\text{v}_{0_y}^2}{2\text{g}}

The initial vertical velocity, v0y, is calculated as

\displaystyle \text{v}_{\text{0y}}=\text{v}_0\sin \theta _0=\left(560.29\:\text{m/s}\right)\sin 45^{\circ} =396.18\:\text{m/s}

Therefore, the maximum height is

\displaystyle \text{h}_{\text{max}}=\frac{\left(396.18\:\text{m/s}\right)^2}{2\left(9.81\:\text{m/s}^2\right)}

\displaystyle \text{h}_{\text{max}}=8000\:\text{m}       ◀

Part C

Consider the following figure

A right triangle is formed with the legs, the horizontal distance and the radius of the earth, and the hypotenuse is the sum of the radius of the earth and the distance d, which is the unknown in this problem. Using Pythagorean Theorem, and converting all units to meters, we have

\displaystyle \text{R}^2+\left(32.0\times 10^3\:\text{m}\right)^2=\left(\text{R}+\text{d}\right)^2

\displaystyle \left(6.37\times 10^6\:\text{m}\right)^2+\left(32.0\times 10^3\:\text{m}\right)^2=\left(6.37\times 10^6+\text{d}\right)^2

\displaystyle \text{d}=\sqrt{\left(6.37\times \:10^6\:\right)^2+\left(32.0\times 10^3\:\right)^2}-6.37\times \:10^6

\displaystyle \text{d}=80.37\:\text{m}       ◀

This error is not significant because it is only about 1% of the maximum height computed in Part B.


College Physics 3.32 – Ranges of projectiles fired at the same initial velocities with different initial angles


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.

Ranges of projectiles fired with the same initial velocities but at different initial angles
Ranges of projectiles fired with the same initial velocities but at different 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 from the result of the calculations, we can say that the numbers in the figure is verified. The very small differences are only due to round-off errors.


College Physics 3.31 – Ranges of projectiles fired at the same angle with different initial velocities


Verify the ranges for the projectiles in Figure 3.40(a) for θ=45º and the given initial velocities.

Projectiles fired at the same angle (45°) with different initial velocities. The initial velocities are 30 m/s, 40 m/s and 50 m/s.
Projectiles fired at the same angle (45°) with different initial velocities.

Solution:

To verify the given values in the figure, we need to solve for individual ranges for the given initial velocities. 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 velocity is 30 m/s, the range is

\displaystyle \text{R}=\frac{\left(30\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=91.74\:\text{m}        ◀

When the initial velocity is 40 m/s, the range is

\displaystyle \text{R}=\frac{\left(40\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=163.10\:\text{m}        ◀

When the initial velocity is 50 m/s, 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}        ◀

Based from the results, we can say that the ranges are approximately equal. The differences are only due to round-off errors.


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.


College Physics 3.19 – Vector Operations


Do Exercise 3.16 again using analytical techniques and change the second leg of the walk to 25.0 m straight south. (This is equivalent to subtracting B from A — that is, finding R’=A – B ) (b) Repeat again, but now you first walk 25.0 m north and then 18.0 m east. (This is equivalent to subtract A from B —that is, to find A=B+C . Is that consistent with your result?)


Solution:

Part A

From the given statement, you first walk 18.0 m straight west and then 25.0 straight south. These vectors are represented by the graph shown below.

The two displacement vectors A and B, and their resultant R directed θ degrees from the x-axis.
The two displacement vectors A and B, and their resultant R directed θ degrees from the x-axis.

To solve for the resultant, we simply need to use the Pythagorean theorem to solve for the hypotenuse of the right triangle formed. That is,

\displaystyle \text{R}=\sqrt{\text{A}^2+\text{B}^2}

\displaystyle \text{R}=\sqrt{\left(18.0\:\text{m}\right)^2+\left(25.0\:\text{m}\right)^2}

\displaystyle \text{R}=30.8\:\text{m}

To solve for the angle, θ, we shall use the tangent function.

\displaystyle \theta =\tan ^{-1}\left(\frac{\text{B}}{\text{A}}\right)

\displaystyle \theta =\tan ^{-1}\left(\frac{25.0\:\text{m}}{18.0\:\text{m}}\right)

\displaystyle \theta =54.25^{\circ}

Thus, the value of 𝜙 can be calculated as

\displaystyle \phi=90^{\circ} -54.25^{\circ}

\displaystyle \phi=35.75^{\circ}

Therefore, the compass direction of the resultant is 35.75° West of South.

Part B

From the statement, you walk 25.0 m north first and then 18.0 m east. This is represented by the figure shown below.

The displacement vectors A and B, and their resultant R directed 𝜙 degrees from the east axis.

So, we have a right triangle with legs 25.0 m and 18.0 m. We are tasked to solve for the value of R, and the angle 𝜙 for the compass direction. The value of R can be solved using the Pythagorean Theorem as in Part A.

\displaystyle \text{R}=\sqrt{\text{A}^2+\text{B}^2}

\displaystyle \text{R}=\sqrt{\left(18.0\:\text{m}\right)^2+\left(25.0\:\text{m}\right)^2}

\displaystyle \text{R}=30.8\:\text{m}

To solve for the angle, θ, we shall use the tangent function.

\displaystyle \theta =\tan ^{-1}\left(\frac{\text{B}}{\text{A}}\right)

\displaystyle \theta =\tan ^{-1}\left(\frac{18.0\:\text{m}}{25.0\:\text{m}}\right)

\displaystyle \theta =35.75^{\circ}

Therefore, the compass direction of the resultant is 35.75° East of North.