ACT: Airiana the Human Arrow

PRINCIPLES: Projectile Motion and Newton' Third Law


- Students will become familiar with - Students will cooperatively build a working putting numbers in a given equation model of "Airiana the Human Arrow" - Students will learn the variables that - Students will make predictions about affect an object' trajectory Airiana's flgiht path using the model - Students will recognize projectile - Students will learn how Newton's

motion and understand the definition Third Law of Physics works with projectiles


The act that we have chosen to teach is entitled "Airiana the Human Arrow." In the actual circus act, a woman named Airiana is shot out of an extremely large human crossbow and subsequently lands in a net over sixty feet away! The Airiana act is similar to that of a human cannonball. In the Airiana act, however, Airiana is exposed when she is fired from the crossbow. Airiana lies on a platform that is then pulled back by bungee chords. When the platform is released, Airiana goes soaring through the air and lands in a suspended net. Every time that the circus changes locations, they must readjust the firing of the human crossbow. If they aim the crossbow too high, Airiana may hit the ceiling, or not reach her destination. If they aim the crossbow too low, she may miss the net. So, how do the technicians know where to set the crossbow; why is Airiana confident that she will reach the net, and stay in the net, without bouncing out? The answers lie behind some important physic principles.


And The


What is Projectile Motion?

Projectile motion is the motion of an object projected into the air near the surface of the earth and subjected to the downward acceleration of gravity. When dealing with projectile motion, one must think in two directions simultaneously, because there are two motions occurring at the same time. The object has independent motion in the horizontal, and the vertical direction. The only thing that the two motions have in common is that they occur at the same time interval.

The actual path of the object is called trajectory. Some examples of projectile motion are a baseball being thrown, a football being kicked, a golf ball being driven, etc.. When shooting a basketball, without even realizing it, you calculate what you need the trajectory to be. Trajectory is influenced by two variables: speed at take off (which itself is affected by variables), and the angle of take off.

Further explanation of the two directions of trajectory

Projectiles move independently in the x and y direction. The x direction is horizontal; x measures the distance of an object, left to right. The y direction is vertical; y measures the height of an object, down and up.

Trajectory equations use x, and y to express these separate directions, so it is important to become familiar with them. When graphing on a Cartesian plane, x and y are also used. The

x-axis is a horizontal line, and the y-axis is a vertical line. The x and y axes are perpendicular to each other. Cartesian coordinates are expressed like this: (x,y).

What is Newton's Third Law of Physics?

The famous physicist Isaac Newton proposed three laws that still govern Physics today. The law that relates to the Ariana act is his third law of motion. This law states that for every force, there is an equal and opposite reaction force. This is shown by the simple equation :

F(1) = F(2)

In this equation, F(1) is the primary force of the object on the second, and F(2) is the reaction force of the second object acting on the first. This law is the rule by which many forces interact in physical terms.

Newton's third law of motion is extremely important to Airiana's act. When she is propelled through the air at astonishing speed, injury and disaster seem imminent. Her force slams into the net, which is what she is relying upon to catch her. The net indeed slows her down and saves her. The flying Airiana is acting as the primary force on the net. The net is acting as the reaction force on Airiana. The two forces balance as the net gives way so Airiana slows down. When the reaction force is done being enacted, Airiana has almost slowed to a halt and can gracefully lower herself along the net to the ground.

What factors affect Airiana's initial speed?

In the human arrow act, Airiana is the object being projected. The initial speed of Airiana is affected by the bungee chords, and Airiana's weight. The bungee chords are a large factor; the force that they send Airiana can be affected by their length, the temperature, and the strength of the chords.

What is air friction?

When Airiana flies, air friction, or drag, also comes into play. An everyday example of air friction is when you stick your hand out a car window while driving. The force that you feel on your hand, that is pushing your hand backwards is air friction. For ease of calculations, we will not take drag into account. Therefore, do not be concerned with possible inconsistencies, from the model launch, to the mathematic prediction.


Sine and cosine are two ratios taken from a triangle. Look at Fig. 1. Fig.1 is a right triangle with sides a, b, and c. The hypotenuse of the triangle in Fig. 1 is side a. This is because side a is opposite the 90 degree angle. The sine of angle is (b/a) and the cosine of angle is (c/a). ( is used to denote the angle which we are looking at the ratio of). As a rule, the sine of a right triangle is the side opposite the angle divided by the hypotenuse. The cosine of a right triangle is the side adjacent to the angle, divided by the hypotenuse. For example, in Fig. 2 the sine of angle is (3/5) and the cosine of angle is (4/5). Mathematically, we express the English phrase "the sine of angle is: " as sin = ? and "the cosine of angle is:" in English translates to cos = ?. Sine and Cosine are ratios used in higher math such as trigonometry.

For ease of solving equations with Airiana, use the sine and cosine ratio chart provided. The answers are rounded to four decimals.




10° .1736 .9848

15° .2588 .9659

20° .3420 .9397

25° .4226 .9063

30° .5000 .8660

35° .5736 .8192

40° .6427 .7660

45° .7071 .7071

50° .7660 .6428

55° .8192 .5736

60° .8660 .5000

65° .9063 .4226

70° .9397 .3420

* Ratios are rounded to four decimal places.




- Discussion of a Cartesian Plane

What do x and y mean?

- Brainstorm types of projectile motion on blackboard (basketball, baseball, etc.)

Organize students into groups, or have them "partner-up." Then allow each group to choose one of the projectile motion brainstorms, and draw the path of the projectile. Have students share their drawings.


How high will Airiana fly?

This problem is asking for the vertical measurement of Airiana's trajectory, when the velocity of y=0. When the velocity of y=0, Airiana is not rising anymore, and Airiana is beginning her descent. As gravity wins the battle over Airiana and begins to pull her down, y will decrease. The equation for this problem is:

y = (initial speed *sin2)2 / (2*gravity)

In this equation, 2 is used to denote the angle measure at which the projectile object is being released. Gravity is equal to 9.8m/s2. However, you cannot always divide the

(initial speed *sin2)2 by 19.6, because sometimes the initial speed is given in feet per second, as opposed to meters per second. In the case of the initial speed being in feet per second, you must multiply 9.8 by 3.28. This is because there are approximately 3.28 feet in every meter.

How far will Airiana travel?

In this problem we must first say, for ease of problem solving that Airiana is landing at the same height (i.e. zero feet, meter, or inches, etc.) that she takes off from. So, this problem is asking for x when y=0, (the height, as opposed to the velocity). Formerly solving for time (t), we have found:

t = (initial speed*sin2) / (2 * gravity)

We use t in our equation for the length, or time of Airiana's flight:

x = (initial speed * cos2) (t)

We have supplied you with the time equation so that if you wish to find the length Airiana's flight, you can do that as well.


e.g. #1 Q: Airiana is shot from the human crossbow at an initial speed of 65 ft/s2 and at an angle of 35°. How far does she travel, and what is the maximum height that she attains?

A: First, determine what gravity is, and the sine and cosine of 35°.

gravity = 9.8m/s2 Multiply gravity by 3.28, because the speed

gravity = 32.144 ft/s2 given in feet

sin 35° = .5736 Use the Sine & Cosine Ratio Table

cos 35° = .8192


y = (initial speed * sin2)2 / (2 * gravity) Model equation (always write out the model equation before plugging in the numbers)

y = (65 * .5736)2 / (2*32.144)

= (1390.0967) / (64.288)

. 21.6230


t = (initial speed * sin2) / (2*gravity) Model equation

t = (65 * .5736) / (2 * 32.144) First solve for the time

= (37.284) / (16.072)

. 2.3198 seconds

x = (initial speed * cos2) (t) Model equation

x = (65 * .8192)(t) Then using the time, solve for the distance

. 123. 5247

MAX HEIGHT : 21.623 feet Final Answer, must include unit of

DISTANCE: 123.5247 feet measurement


adjacent: next to; touching.

Cartesian Plane: An (x,y) coordinate system.

force: A push or pull.

horizontal: The x direction of a graph. Horizontal measure is distance, and goes from left to right. (Mnemonic: Horizontal comes from horizon)

hypotenuse: In a right triangle, the side opposite the right angle is the hypotenuse of the triangle.

perpendicular: A ninety degree angle formed by two lines.

projectile motion: The motion of an object from the combination of movement in the x and y direction.

trajectory: The path of a projectile.

vertical: The y direction of a graph. Vertical measure is height, and goes high to low.





1. Make the model! (Instructions on p. #)

2. Crossword Puzzle

3. Word Find

*4. Projectile Problem worksheet ansd/or have students write their own word problem about Airiana's trajectory

5. Class Sing-a-Long to "Dream of Flight" song (create your own tune, lyrics on next page)

6. Have students write down five examples of projectile motion.

7. Research Isaac Newton and/or his other laws of motion.

8. Have students answer the following essay questions.

1) How does Newton's Third Law affect Ariana's flight? Why? (Sample Answer)

Newton's third law is the relationship of equality between primary force and reaction force. Ariana is slowed down at the end of her flight by the net positioned precisely where she lands. The net's reaction force nullifies Ariana's primary force, thus slowing her down.

2) Explain exactly how the five specific examples discovered for homework incorporate projectile motion principles. (Extra credit if Newton's Third Law is mentioned!)

Answers will vary.

* Only proceed with these activities if you feel that your students are easily grasping the projectile motion concepts. Depending on the prior knowledge of students, they may not be ready to do the math necessary in the problems provided.

Someone is waiting to thrill you,

Fulfill all your fantasies.

Gravity is hers to defy.

Keep your eye on the sky!


She has found a way to inspire.

She has touched the fire of the sun!

Haven't you ever wished you could fly?

Keep your eye on the sky!


The bravest of the brave will astound you with her beauty.

Can you feel all the courage that she brings?

The hero of our day ascends like an explosion

And proves that we can fly without wings!

Fly brave hero, like a shooting star!

Human Arrow, show us who you are!

Fly, earth angel, teach us how it's done!

Airiana, shining like the sun!

Glorious! Victorious! Glowing heroine!

Legend of the sky! Weightless as the wind!



(1) Airiana is shot from the human crossbow at an angle of 45°, and an initial speed of 20 m/s2. Will she land in a net ten feet long that begins 54 meters away?

(2) Airiana is shot from the human crossbow at an angle of 40° and an initial speed of 30m/s2. What would be the difference in Airiana's maximum height if the angle is lowered to 30°?

(3) Airiana is shot from the human crossbow at an angle of 45°, and an initial speed of 40 ft/s2. What is her maximum height, and how long is she in the air?

(4) Airiana is shot from the human crossbow at an angle of 30° and an initial speed of 18m/s2. Will she hit a ceiling that is 8 meters high? How far away should Airiana's landing net be placed?


(1) This problem requires that you first find the distance covered by Airiana's flight. First, determine the sine and cosine of 45°, and then the force of gravity.

gravity = 9.8m/s2 sin 45° = .7071 cos 45° = .7071

Then solve for time:

t = (initial speed * sin2) / (2 * gravity) Model equation for time

t = (20 * .7071) / (2 * 9.8)

= (14.142) / (4.9)

= 2.886

Then solve for the distance:

x = (initial speed * cos2) (t) Model equation for distance

x = (20 * .7071)(t)

= 40.8138

Since 40.8138 is less than 54, Airiana will not land in the net. Her initial speed must be increased, else the net needs to be moved closer.

(2) This problem requires that you solve for the height of the 40° launch, and the height of the 30° launch. Then the 30° launch height must be subtracted from the 40° launch height.

Begin by writing gravity, the sin40°, and the sin30°.

gravity = 9.8 m/s2 sin40° = .6427 sin30° = .5000

Then solve for Airiana's maximum height from the 40° launch:

y = (initial speed * sin2)2 / (2 * gravity) Model equation for height

y = (30 * .6427)2 / (2* 9.8) Numbers from the 40° launch

= (371.7596) / (19.6)

= 18.9671 Answer from the 40° launch

Then solve for Airiana' maximum height from the 30° launch:

y = (30 * .5)2 / (2 * 9.8) Numbers from the 30° launch

= (225) / (19.6)

= 11.4796 Answer from the 30° launch

18.9671 - 11.4796 = 7.4875 Subtract height of 30° launch from height

of 40° launch

ANSWER: The difference in Airiana' maximum height is 7.4875 meters


(3) This problem first asks you for Airiana' maximum height, and then the time, or length of her flight. Begin by writing gravity, and the sin45°.

gravity = 9.8m/s2 = 32.144 ft/s2 sin45° = .7071

Then solve for Airiana' maximum height:


y = (40 * .7071)2 / (2 * 32.144)

= (799.9846) / (64.288)

= 12.4437

Now we solve the length, or time elapsed of Airiana' flight:


t = (40 * .7071) / (2 * 32.144)

= (28.284) / (16.072)

= 1.7598

ANSWER: Airiana' highest point is 12.4437 feet, and she is in the air for 1.758 seconds

(4) This problem requires that you find Airiana' maximum height, and the distance of her flight. Begin by writing gravity, and the sine and cosine of 30°.

gravity = 9.8m/s2 sin35° = .5000 cos30° = .8660

Then solve for Airian's maximum height:


y = (18 * .5000)2 / (2 * 9.8)

= (81) / (19.6)

= 4.1326

Then solve for the distance of Airiana' flight: (begin solving for time)


t = (18 * .50) / (2 * 9.8)

= (9) / (4.9)

= 1.8367

Then solve for the distance:


x = (18 * .8660) (1.8367)

= 28.6304

ANSWER: Ariana will not hit a ceiling that is 8 meters high. Airiana's landing net should be placed 25-30 meters away.