PROJECTILE MOTION

I. Definitions

• A. Projectile Motion – refers to the motion of an object that is projected into the air at an angle. (That angle can be zero!) We assume this occurs near the earth’s surface so “g” is constant at -9.81 m/s2.

• We are NOT concerned with the process of throwing or how it’s projected or what happens afterwards

Trajectory• B. Trajectory

– the name we give the path a projectile follows

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II. Background

• First described by Galileo, who said…

• “An object projected horizontally will reach the ground at the same time as an object dropped vertically”

• We must analyze 2 components of motion separately:

• A) horizontal (the “x” component)

• B) vertical (the “y” component)

DEMO – “Newton’s Law Apparatus”A spring-tension bar is held back by a lever. When the lever is pushed, the bar moves forward. Sphere “A” falls at the same moment sphere “B” is projected forward.

A BA B

Demo – Motion in 2 Dimensions• We can give the sphere an even greater vx…

This green ball is dropped…

…at the same time this yellow ball is “flicked”...

…and at the same time this purple ball is shot.

Component VectorsThe first ball has only a vertical velocity component

The second ball has a small vx and the same vy

as the green ball

The third ball has a larger vx and the same vy as the other two balls

Notice that the instantaneous vy is the same for each ball

Notice that vx for a ball is the same throughout

One more thing…the red line here shows the vector RESULTANT of the blue and green lines.

Solving Projectile Motion Problems

• Analyze horizontal and vertical components separately

• vx never changes throughout the trajectory

• These are generally 2-part problems

Did you catch that? vx never changes throughout the trajectory!

Example

• A rock is kicked horizontally from a 100. m high cliff. It strikes the level ground 90.0 m from the base of the cliff. At what speed was it thrown?

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Problem Solving

• We’re really solving for vx, but vx is in a formula dx = vxt and we don’t know t.

• First, let’s solve for t

• That will tell us the time it takes for the object to fall

How Long Does It Take to Fall?

• Falling time depends on falling distance• Use a formula that solves for time and uses

displacement and gravity• d = ½ at2 but since this is motion in 2-d, let’s call this

displacement a dy.• dy = ½ at2; 100. = ½ (9.81) (t2)• t2 = 20.4; t = 4.52s

What is the initial velocity?

• Initial velocity is v in the x direction

• Now we can solve for vx

• dx = vxt

• 90.0 = v (4.52)

• v = 19.9 m/sWhat is true about v in the x direction each moment that this object is falling?

Now, work in small quiet groups on the problems on side 1 of the

handout.

Part 2 – Projectiles launched at an angle other than zero

• The previous section addressed projectiles launched at zero (horizontally)

• When projectiles are launched at an angle, you must use trigonometry (soh cah toa) to resolve for the component vectors that make up the resulting movement.

Visualizing Projectile Motion

Note that although there is no vy at the top of the path, there is vx and the object is continually affected by ag.

Tips

• If you ever see a projectile problem that asks…

• What is the v at the projectile’s maximum height?

• The answer will always be the same as your answer for vx since @ max. height there is no vy, only vx.

Concept

• Think of projectile motion of objects launched at an angle as an infinite number of vector diagrams with their resultants adding together to form an overall trajectory.

Projectiles launched at an angle

Though the object does not follow this path, the mathematical relationships are the same

Example

• A projectile is fired at an angle of 30.0o with the horizontal at a velocity of 39.2 m/s.

• First, solve for vy and vx

Solve for vy

• vy = sin = opp/hyp• sin 30.0o = vy / 39.2• 0.500 = vy / 39.2• vy = 19.6 m/s

Solve for vx

• vx = cos = adj / hyp;

• cos 30.0o = vx / 39.2;

• 0.866 = vx / 39.2

• vx = 33.9 m/s

What is the max. height reached?

• Max height = dy

• You’ll be using the formula: dy = vyt + ½ at2

• But you must first find t

• t = -2vy / a (Important formula!)

• t = -2 (19.6) / -9.81; t = 4.00s• But that represents the total

time and the max. height occurs at the midpoint, which occurs at ½ the total time!

• (so divide the time by 2!)

dy

But we still haven’t finished!

Find the Maximum Height• Max. height can be called dy

• dy = vyt + ½ at2 you don’t use 4.00s, use 2.00s

• dy = 19.6(2.00) + (-4.91)(2.00)2 = 19.6m

• Careful with the signs! “a” must be -9.81 m/s2 or you’ll be quite a bit off.

• One more question…

What is the maximum range?• Range of course is dx

• dx = vxt

• dx = 33.9 (4.00)

• (must use 4.00 since the range is achieved after the whole trip has been made)

• dx = 136m

•Now, work in small quiet groups to solve the problems on “Projectile Motion – Launched at an Angle”.