Projectile Motion: Questions to Ponder

• What are the three factors that will determine how far an object will travel when thrown?

• What are the two things a vector must show?

• Why does a baseball pitcher throw from a

mound?

Analyzing Projectile Motion

Projectile motion is a motion resultingfrom an initial velocity and subject only

to the acceleration of gravity.

Analyzing Projectile Motion

The time it takes for anobject to reach its apex is the same as the time it takes to go from the apex to the ground.

3 seconds up 3 seconds down

Analyzing Projectile Motion

The initial velocity is the same as the final velocity. Initial velocity is a POSITIVE number; final velocity is NEGATIVE.

3 seconds up

3 seconds down

Shoot the monkey (with a banana…)

StM – with gravity

StM – aim at monkey

StM – slo-mo

Characteristics of Projectile• Horizontal & vertical components are

independent.

• Must be examined separately

• Only vertical force (g) at work

Acceleration• There are NO horizontal forces acting upon

projectiles and thus NO horizontal acceleration

• The horizontal velocity of a projectile is constant (a never changing in value)

• There is a vertical acceleration caused by gravity; its value is 9.8 m/s/s, down = -9.8m/s/s

Cannonball – Vi = 20 m/s

Projectile MotionLaws of Constant Acceleration

Both dimensions follow the kinematic motion equations we used before

Vf=Vi + at

d=1/2(Vf + Vi)t

d=Vit + 1/2at2

Vf2 = Vi

2 + 2ad

Vertical Component

Horizontal

Component

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Did you miss us?

Horizontally launched projectile problems – part x

• Equations for horizontal motion:

But wait - ax = 0 m/s2 …

Horizontally launched projectile problems – part y

• Equations for vertical motion:

Suggestions for Solving PM Problems

1. Visualize the problem – make a drawing2. Decide what your coordinate system is: which directions are +,which

are –3. Remember the time variable, t, is what binds the x motion to the y

motion (same value for the part of the motion).4. Write down all your known and implied values in an organized way5. Verify that the given information contains values for at least 3

of the kinematic variables. Do this for both x and y motions.Then select the appropriate equation(s).

6. You may want to divide the motion into segments. Rememberthe final velocity for one segment is the initial velocity for thenext segment.

7. Remember that a problem may have two possible answers. Tryto visualize the different physical situations to which theanswers correspond.

Variablestime t t acceleration a  ax

  ay

initial velocity vi vi,x

  vi,y

final velocity vf vf,x

  vf, y

displacement d x  y

Example – Launched at Zero Angle

115 m/s

A plane flying at an altitude of 1050 m and a speed of 115 m/splans to drop a bunker-buster bomb on (insert bad guys names

here)’s hideout.How long will it take the bomb to reach its target?How fast will the bomb be traveling upon impact (net)?How far (surface distance) from the target should the bomb be released?

How many questions are here?

Find your variables and then take it one question at a time.

Breathe – YOU CAN DO THIS! 1050 m

Example – Launched at Zero Angle

115 m/s

A plane flying at an altitude of 1050 m and a speed of 115 m/splans to drop a bunker-buster bomb on (bad guy)’s hideout.How long will it take the bomb to reach its target?

1050 m

Fall time is independent of the horizontal speed

Y-dimension values:

Vi = 0 a= 9.80 m/s2 d = 1050 m t = ?

apply: d=Vit + 1/2at2 , 2d/a = t2 , √2d/a = t

t = √2(1050m)9.80m/s2

t = 14.6s

Example – Launched at Zero Angle

115 m/s

A plane flying at an altitude of 1050 m and a speed of 115 m/splans to drop a bunker-buster bomb on (bad guy)’s hideout.How fast will the bomb be traveling upon impact (net)?

1050 m

115 m/s

? m/s net

The vertical velocity upon mpact is independent of the horizontal speed

Vi = 0 a= 9.80 m/s2 d = 1050 m Vf = ?

Apply: Vf2 = Vi2 + 2ad , Vf = √Vi2 + 2ad

Vf = √2(9.80m/s2)1050m

Vf = 144 m/s

Example – Launched at Zero Angle

115 m/s

A plane flying at an altitude of 1050 m and a speed of 115 m/splans to drop a bunker-buster bomb on (bad guy)’s hideout.How fast will the bomb be traveling upon impact (net)?

1050 m

115 m/s

144 m/s net

Use the Pythagorean theorem to resolve two vectors occurring at right angles.

C2 = A2+ B2

C = √A2+ B2

C = √(144 m/s)2 + (115 m/s)2

C = 184

Example – Launched at Zero Angle

115 m/s

A plane flying at an altitude of 1050 m and a speed of 115 m/splans to drop a bunker-buster bomb on (bad guy)’s hideout.How far (surface distance) from the target should the bomb be released?

1050 m

The horizontal motion is independent of the vertical motion… except that the time that the projectile is in the air MUST be the same for both dimensions. The horizontal frame of reference is not accelerating therefore…

d = v (t) d = 115m/s (14.6s)

d = 1680 m

YOU TRY

• PG. 102 #1-4

What if the object is projected at an angle?

• Then we can use the following equations:» Vx=Vi(cosΘ)

»Δx=Vi(cos Θ) Δt

»Vyf2=Vi

2(sin Θ)2 + 2aΔy

»Δy=Vi(sin Θ) Δt + 1/2a(Δt)2

» Vyf=Vi(sin Θ) + aΔt

• PG. 104 #1-5