bk 2 ch 6 projectile motion

Unit 6 Projectile Motion

Projectiles, Trajectory and Inertia

The most common example of an object that is moving in two dimensions is a

projectile. A projectile is an object upon which the only force acting is

gravity. There are a variety of examples of projectiles. Any thrown object is a

projectile, assuming that air resistance is negligible. A projectile is any object

that once projected or dropped continues in motion by

its own inertia and is influenced only by the downward

force of gravity.

By definition, a projectile has a single force that acts

upon it - the force of gravity. If there were any other

force acting upon an object, then that object would not

be a projectile. Thus, the free-body diagram of a projectile would show a single

force acting downwards and labeled force of gravity, regardless of whether a

projectile is moving in any direction.

Many students have difficulty with the concept that the only force acting

upon an upward moving projectile is gravity. Their conception of motion

prompts them to think that if an object is moving upward and rightward,

there must be both an upward and rightward force. Their belief is that

forces cause motion. Such

students do not believe

strongly in Newtonian physics.

Newton's laws suggest that

forces are only required to

cause an acceleration. A

force is not required to keep

an object in motion, as stated

in Newton’s first law. Consider

a cannonball shot horizontally

from a very high cliff at a high speed. And suppose for a moment that the

gravity switch could be turned off such that the cannonball would travel in the

absence of gravity? What would the motion of such a cannonball be like? How

could its motion be described? According to Newton's first law of motion, such

a cannonball would continue in motion in a straight line at constant speed. If

not acted upon by an unbalanced force, "an object in motion will ..."

Gravity acts to influence the vertical motion of the projectile, thus causing a

vertical acceleration. Due to the absence of horizontal forces, a projectile

remains in motion with a constant horizontal velocity !

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Learn more : http://www.physicsclassroom.com/mmedia/vectors/mzng.cfm

In projectile motion, vertical and horizontal

components are discussed separately. Using the

cannon on a tall cliff as a guiding example, the

projectile travels with a constant horizontal

velocity and a downward vertical acceleration

when gravity is switched on. The ball’s

acceleration is 9.8 downwards and its velocity

(combined components) is therefore changing.


Learn more : http://www.physicsclassroom.com/mmedia/vectors/hlp.cfm

Learn more (Launched at an angle) :

http://www.physicsclassroom.com/mmedia/vectors/nhlp.cfm

Checkpoint 1

1.Suppose a snowmobile is equipped with a flare launcher that is

capable of launching a sphere vertically (relative to the

snowmobile). If the snowmobile is in motion and launches the

flare and maintains a constant horizontal velocity after the

launch, then where will the flare land (neglect air resistance)?

___________________________________

2.Suppose a rescue airplane drops a relief package while it is

moving with a constant horizontal speed at an elevated height.

Assuming that air resistance is negligible, where will the relief

package land relative to the plane? ____________________________

Learn more for Question 1 and 2 :

http://www.physicsclassroom.com/mmedia/vectors/tb.cfm

http://www.physicsclassroom.com/mmedia/vectors/pap.cfm

Calculating Components for Simple Projectile Motion

Consider again the cannonball

launched by a cannon from the top of

a very high cliff. Suppose that the

cannonball is launched horizontally

with no upward angle whatsoever

and with an initial speed of 20 m/s. If

there were no gravity, the cannonball

would continue in motion at 20 m/s in

the horizontal direction. Yet in

actuality, gravity causes the

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http://www.physicsclassroom.com/mmedia/vectors/mzng.cfm

cannonball to accelerate downwards at a rate of 9.81 m s-2. This means that the

vertical velocity is changing by 9.81 m/s every second.

Similarly, if it is assumed that an object is projected at an angle upwards with

a velocity of 19.6 m/s, then after 1 sec it becomes 9.8 m/s, then after another

second it is momentarily at rest in midair, and falls later.

Now let us discuss calculations related to horizontal and vertical components

of projectiles. Take Y as the vertical displacement and X as the horizontal

displacement. Y can be calculated using the equation learnt from basic

kinematics :

, where g = acceleration due to gravity and t = time in

seconds

Since the horizontal component is only influenced by the initial velocity, X

can be calculated by : , where v = initial velocity and t = time in seconds

For calculating Y in situations where the projectile is launched at an angle

to the horizontal :

, where u is the initial velocity of the projectile.

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Checkpoint 2

1.Annaline drops a ball from rest from the top of 78.4-meter high

cliff. How much time will it take for the ball to reach the ground

and at what height will the ball be after each second of

motion?

2.A cannonball is launched horizontally from the top of an 78.4-

meter high cliff with a velocity of 4 m/s. How much time will it

take for the ball to reach the ground and at what height will the

ball be after each second of travel?

3.Refer to Question 2, draw a plot diagram to represent the

trajectory of the ball. Hence, calculate the displacement of the

ball in vertical two-dimension.

4.A cannonball is projected from a tall cliff of 1000m with an

initial velocity. After certain time, it falls onto the ground

1000m below the cliff, 5000m away from the initial position in

horizontal direction. Find the magnitude of the initial velocity

Answers for Checkpoint 1 and Checkpoint 2

1. The flare lands inside the the snowmobile 2. The package lands directly

below the plane

Checkpoint 2 :

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1. 4 seconds 2. 4 seconds 3. 80m, S11.5 E

4.350 m/s

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Initial Velocity Components

We had just discussed that the horizontal and vertical motions of a projectile

are independent of each other. Thus, an analysis of the motion of a projectile

demands that the two components of motion are analyzed independent of

each other, being careful not to mix horizontal motion information with

vertical motion information. Vector resolution is the method of taking a single

vector at an angle and separating it into two perpendicular parts. The analysis

of projectile motion problems begins by using trigonometric methods to

determine the horizontal and vertical components of the initial velocity.

Practice vector resolution with the following examples by resolving the initial

velocity !

1.A water balloon is launched with a speed of 40 m/s at an angle

of 60 degrees to the horizontal

2.A motorcycle stunt person traveling 31.3 m/s jumps off a ramp

at an angle of to the horizontal

3.A springboard diver jumps with a velocity of 10 m/s at an angle

of 80 degrees to the horizontal.

Suggested answers in corresponding order :

>Horizontal components : 20, 25.6, 1.7 >Vertical components : 34.6,

17.9, 9.8

After understanding how to resolve velocities into components, you can

pretty-much solve any problems involving projectile motion! Let us look at the

first type of problem, which is finding the time of flight of a projectile :

The time for a projectile to rise vertically to its peak (as well as the time to

fall from the peak) is dependent upon vertical motion parameters. The process

of rising vertically to the peak of a trajectory is a vertical motion and is thus

dependent upon the initial vertical velocity and the vertical acceleration (g =

9.8 m/s2, down).

For example, if a projectile is moving upwards with a velocity of 39.2 m/s at 0

seconds, then its velocity will be 29.4 m/s after 1 second, 19.6 m/s after 2

seconds, and finally 0 m/s after 4 seconds. It takes 4 seconds for it to reach the

peak where its vertical velocity is 0 m/s. With this notion in mind, it is evident

that the time for a projectile to rise to its peak is a matter of dividing the

vertical component of the initial velocity by the acceleration of gravity :

Another problem is the determination of the peak height achieved by the

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projectile. We can simply use the formula for calculating vertical displacement

Y when launched at an angle, and substitute t as time required to reach the

peak.

Checkpoint 3

1.Bob Beamon has made a record-breaking long jump in the

Olympic games in 1986. He jumped a horizontal distance of 8.9

m with an initial velocity of 9.5 m/s at an angle of .

a)It is known that the runway before his leap is 30 m long.

Calculate Bob’s average acceleration before his jump.

b) Find the peak height reached by Bob Beamon. Hence,

determine whether the same jump can be used to leap over a

high-jump fence of 2 m height from the ground.


2. Megan is a golf pro. She hits a ball with a velocity of 25 m/s at an angle of to the horizontal

a)Megan’s golf club has a mass of 2 kg and its velocity before

hitting the ball is 12 m/s. After it hits the ball, it stops

immediately. Determine the mass of the ball.

b) Determine the time of flight of the golf ball if the landing

point of the ball is at the same level as the original position of

the ball.

c)Megan wants the golf ball to fly for at least 5 seconds before

landing. Determine the minimum initial velocity required to

achieve this

3. Kevin launches a football at an angle of to the horizontal. The football

reaches a horizontal distance of 79.5 m when it first lands at the same level.

a)Determine the initial velocity of the football (Hint : Set up two

equations)

b) Kevin wants to launch the football to half the distance with

the same initial velocity. What is the angle to the horizontal

needed to achieve this?

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4.Fill in the blanks in the table below

A = __________

B = __________

C = __________

D = __________

E = __________

F = __________

G = __________

H = _________ I = __________ J = __________ K = __________

L = _________ M = __________ N = __________ O = ___________

P = _________ Q = __________ R = __________

Answers for Checkpoint 3

1a) 1.50 m/s2 1b) 1.9 m, He cannot 2a) 1 kg 2b) 4.42 second

2c) 49 m/s 3a) 30 m/s 3b) 4) A: 14.9 m B: 164

m

C: 2.93 s D: 5.85 s E: 42.0 m F: 240 m G:

19.2 m/s

H: 16.1 m/s I: 1.64 s J: 3.28 s K: 7 L:

19.7 m/s

M: 2.01 s N: 4.02 s O: 19.7 m/s P: 3.47 m/s Q: 0.354 s R:

0.709 s


Mathematical and Algebraic Problems in Projectile Motion

Let’s look at some arithmetic real-life applications of calculations related to

projectile motion. You can practice using the example questions but there is no

answers for reference!

Type 1 : Falls off an edge – A projectile is launched with an initial horizontal

velocity from an elevated position and follows a parabolic path to the ground.

Questions may ask initial speed, initial height, time of flight, horizontal

distance of the projectile

E.g. (1) : A pool ball leaves a 0.60-meter high table with an initial horizontal

velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the

ground and the horizontal distance between the table's edge and the ball's

landing location

Type 2 : Parabolic Catapult – A projectile is launched at an angle to the

horizontal and rises upwards to a peak while moving horizontally. Upon

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reaching the peak, the projectile falls with a motion that is symmetrical to its

path upwards to the peak. Questions may ask time of flight, horizontal range,

and peak height

E.g. (2) : A football is kicked with an initial velocity of 25 m/s at an angle of 45-

degrees with the horizontal. Determine the time of flight, the horizontal

distance, and the peak height of the football

Remember to make good use of kinematic equations to solve horizontal

or vertical projectile motion. Those equations are :

> > >

Below is an example which shows how to solve a projectile problem :

Q : A pool ball leaves a 0.60-meter high table with an initial horizontal velocity

of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and

the horizontal distance between the table's edge and the ball's landing

location.

This type of question simply requires you to choose suitable kinematic

equations for calculation. Type 2 questions which are more complicated can

ask you to set up 2 simultaneous equations with 2 unknowns, from the

horizontal and vertical dimensions of the projectile’s motion respectively.

A : Time required is 0.350 s and horizontal distance is 0.84 m

Try these questions ! Only one answer is provided for the last question,

though !

1.A soccer ball is kicked horizontally off a 22.0-meter high hill

and lands a distance of 35.0 meters from the edge of the hill.

Determine the initial horizontal velocity of the soccer ball.

2.A long jumper leaves the ground with an initial velocity of 12

m/s at an angle of 28-degrees above the horizontal. Determine

the time of flight, the horizontal distance, and the peak height

of the long-jumper.

3.A soccer ball is kicked horizontally off a 22.0-meter high hill

and lands a distance of 35.0 meters from the edge of the hill.

Determine the initial horizontal velocity of the soccer ball.

Answers for Q3 = 16.5 m/s


Checkpoint 4

1. Determine the launch speed of a horizontally launched projectile that lands

26.3 meters from the

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base of a 19.3-meter high cliff

2. A soccer ball is kicked horizontally at 15.8 m/s off the top of a field house

and lands 33.9 metes from the base of the field house. Determine the height of

the field house

3a) Determine the range of a ball launched with a speed of 40.0 m/s at angles

of … from ground level.

(i) 40.0 degrees

(ii) 45.0 degrees

(iii) 50.0 degrees

3b) For the three initial launch angles in question 3a, determine the

corresponding peak heights. Hence, deduce which angle can launch the ball

the highest

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3c) For the three initial launch angles in question 3a, determine the

corresponding horizontal distances. Hence, deduce which angle can launch the

ball the furthest

3. A zookeeper has a monkey that he must feed daily. The monkey spends most

of the day in the trees

just hanging from a branch. When the zookeeper launches a banana to the

monkey, the monkey has

the peculiar habit of dropping from the trees the moment that the banana is

launched.

The banana is launched with a speed of 16.0 m/s at a direction of 51.3° above

the horizontal (which

would be directly at the monkey). The monkey is initially at rest in a tree 25.0-

m above

the

ground.

Determine the horizontal and

vertical displacements of the

banana and the monkey at 0.5-

second time intervals. Then plot

the trajectories of both banana

and monkey on the diagram

below.

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bk 2 ch 6 projectile motion

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