Freely Falling ObjectsNear the surface of the Earth, all objects experience approximately the same acceleration due to gravity.

This is one of the most commonThis is one of the most commonexamples of motion with constantacceleration.

Figure : Multiflash photograph of afalling apple, at equal timeintervals. The apple falls fartherduring each successive interval,g ,which means it is accelerating.

In the absence of airIn the absence of airresistance, all objects fallwith the same acceleration,

lth h thi b t i kalthough this may be trickyto tell by testing in anenvironment where there isair resistance.

Figure: (a) A ball and a light piece of paper are dropped at the same time. (b) Repeated, with the paper wadded up

The acceleration due to gravityg yat the Earth’s surface isapproximately 9.80 m/s2. At agiven location on the Earth andgiven location on the Earth andin the absence of air resistance,all objects fall with the same

t t l ticonstant acceleration.

Figure : A rock and a feather are dropped simultaneously (a) in air, (b) in a vacuum.

Example 1: Falling from a tower.

S h b ll i d dSuppose that a ball is dropped(v0 = 0) from a tower 70.0 m high.How far will it have fallen after atime t1 = 1.00 s, t2 = 2.00 s, and t3= 3.00 s? Ignore air resistance.

Figure: (a) An object dropped from atower falls with progressively greaterspeed and covers greater distance witheach successive secondeach successive second.(b) Graph of y vs. t.Solution: We are given the acceleration,the initial speed and the time; we needthe initial speed, and the time; we needto find the distance. Substituting gives x1= 4.90 m, x2 = 19.6 m, and x3 = 44.1 m.

Example 2 : Thrown down from a tower.p

Suppose a ball is thrown downward with an initial velocity of3.00 m/s, instead of being dropped. (a) What then would beits position after 1.00 s and 2.00 s? (b) What would its speedbe after 1.00 s and 2.00 s? Compare with the speeds of adropped balldropped ball.

Solution: This is the same as Example 1, except that the initialspeed is not zero.

a.At t = 1.00 s, y = 7.90 m. At t = 2.00 s, y = 25.6 m.b. At t = 1.00 s, v = 12.8 m/s. At t = 2.00 s, v = 22.6 m/s. The

speed is always 3 00 m/s faster than a dropped ballspeed is always 3.00 m/s faster than a dropped ball.

E l B ll th d IExample : Ball thrown upward, I.

A person throws a ball upward into the air with aninitial velocity of 15.0 m/s. Calculate (a) how high itgoes, and (b) how long the ball is in the air before itcomes back to the hand. Ignore air resistance.

Figure: An object thrown into the air leaves thethrower’s hand at A, reaches its maximum height at B,and returns to the original position at C.Solution: a At the highest position the speed is zeroSolution: a. At the highest position, the speed is zero,so we know the acceleration, the initial and finalspeeds, and are asked for the distance. Substitutinggives y = 11 5 mgives y = 11.5 m.b. Now we want the time; t = 3.06 s.

Conceptual Example : Two possible misconceptions.

Give examples to show the error in these two commoni ti (1) th t l ti d l itmisconceptions: (1) that acceleration and velocity are

always in the same direction, and (2) that an object thrownupward has zero acceleration at the highest point.g

1.If acceleration and velocity were always in the samey ydirection, nothing could ever slow down!

2. At its highest point, the speed of thrown object is zero. Ifits acceleration were also zero it would just stay at thatits acceleration were also zero, it would just stay at thatpoint.

Example : Ball thrown upward, II.

L id i b ll hLet us consider again a ball thrownupward, and make more calculations.Calculate (a) how much time it takes for( )the ball to reach the maximum height, and(b) the velocity of the ball when it returns tothe thrower’s hand (point C)the thrower s hand (point C).

a.The time is 1.53 s, half the time for a roundtrip (since we are ignoring air resistance).trip (since we are ignoring air resistance).

b. v = -15.0 m/s

Example : Ball thrown upward, III; the quadratic formula.

For a ball thrown upward at an initial speed of 15.0 m/s, calculate atwhat time t the ball passes a point 8.00 m above the person’s hand.

Figure : Graphs of (a) y vs. t, (b) v vs. t for a ball thrown upward, Solution: We are given the initial and final position, the initial speed, and thegacceleration, and want to find the time. This is a quadratic equation; there aretwo solutions: t = 0.69 s and t = 2.37 s. The first is the ball going up and thesecond is the ball coming back down.

Ball thrown upward at edge of cliff.

S th t b ll i th dSuppose that a ball is thrown upwardat a speed of 15.0 m/s by a personstanding on the edge of a cliff, so thatg gthe ball can fall to the base of the cliff50.0 m below. (a) How long does ittake the ball to reach the base of thetake the ball to reach the base of thecliff? (b) What is the total distancetraveled by the ball? Ignore air

i t (lik l t b i ifi tresistance (likely to be significant, soour result is an approximation).

Figure: The person in Fig. stands on the edge of a cliff. The ball falls to thebase of the cliff, 50.0 m below.Solution: a. We use the same quadratic formula as before, we find t = 5.07 sq ,(the negative solution is physically meaningless).b. The ball goes up 11.5 m, then down 11.5 m + 50 m, for a total distance of73.0 m.

Projectile Motion

A projectile is an object moving in two dimensions underthe influence of Earth's gravity; its path is a parabola.

It can be understood byl i th h i t l danalyzing the horizontal and

vertical motions separately.

The speed in the x‐direction isconstant; in the y‐direction the object

ith t t l timoves with constant acceleration g.

This photograph shows two balls thatstart to fall at the same time The onestart to fall at the same time. The oneon the right has an initial speed in thex‐direction. It can be seen that verticalpositions of the two balls are identicalat identical times, while the horizontalposition of the yellow ball increasesp ylinearly.

If an object is launched at an initial angle of θ0 with thehorizontal the analysis is similar except that the initialhorizontal, the analysis is similar except that the initialvelocity has a vertical component.

The motion of an object in a vertical plane under the influence of it ti l f i k “ j til ti ”gravitational force is known as “projectile motion”

The projectile is launched with an initial velocity ov

The horizontal and vertical velocity components are:

cosv v θ= sinoy o ov v θ=cosox o ov v θ oy o o

Projectile motion will be analyzed in

g

Projectile motion will be analyzed ina horizontal and a vertical motionalong the x- and y-axes,respectively These two motions arerespectively. These two motions areindependent of each other. Motionalong the x-axis has zeroacceleration. Motion along the y-axisg yhas uniform acceleration ay = -g

( ) ( )2

0 0

cos (eqs 2) sin (eqs 4)

The equation of the pa :thgtx v t y v tθ θ= = −( ) ( )

( ) 2

0 0 cos (eqs.2) sin (eqs.4)2

If we eliminate between equations 2 and 4 we get:

Thi i d

oo

g

x v t y v

t

t

θ

θ θ

ib h h f h i( )( )

22tan

2 cos This equatio e dno

o o

gy x xv

θθ

= −

2

scribes the path of the motion

The path equations has the form: This is the equation of a parabolay ax bx= +

The equation of the path seems toocomplicated to be useful Appearances canNote:complicated to be useful. Appearances candeceive: Complicated as it is, this equationcan be used as a short cut in many projectilemotion problems

Solving Problems Involving Projectile Motion

Projectile motion is motion with constant acceleration intwo dimensions, where the acceleration is g and is down.

Example : Driving off a cliff.

A movie stunt driver on a motorcycle speeds horizontally off a 50 0‐m‐A movie stunt driver on a motorcycle speeds horizontally off a 50.0 mhigh cliff. How fast must the motorcycle leave the cliff top to land onlevel ground below, 90.0 m from the base of the cliff where the

? I i i tcameras are? Ignore air resistance.

Example : A kicked football.

A football is kicked at an angle θ = 37 0° with a velocity of 20 0 m/sA football is kicked at an angle θ0 = 37.0 with a velocity of 20.0 m/s,as shown. Calculate (a) the maximum height, (b) the time of travelbefore the football hits the ground, (c) how far away it hits theground, (d) the velocity vector at the maximum height, and (e) theacceleration vector at maximum height. Assume the ball leaves thefoot at ground level, and ignore air resistance and rotation of theg , gball.

A child sits upright in a wagon which is moving to the right at

Conceptual Example : Where does the apple land?

p g g g gconstant speed as shown. The child extends her hand and throws anapple straight upward (from her own point of view), while thewagon continues to travel forward at constant speed If airwagon continues to travel forward at constant speed. If airresistance is neglected, will the apple land (a) behind the wagon, (b)in the wagon, or (c) in front of the wagon?

-Apple has Vxo equal to speed of wagon VxoA l ill f ll th th f j til-Apple will follow the path of projectile as

shown in figure.-Apple experiences no horizontal acceleration so Vxo will stay constant & equal to speed of y q pwagon.-Wagon will be directly under the apple, it will drop into the wagon.

Conceptual Example : The wrong strategy.A boy on a small hill aims his water‐balloon slingshot horizontally, straight at a

d b h i f t b h di t d At th i t t thsecond boy hanging from a tree branch a distance d away. At the instant thewater balloon is released, the second boy lets go and falls from the tree,hoping to avoid being hit. Show that he made the wrong move. (He hadn’tstudied physics yet ) Ignore air resistancestudied physics yet.) Ignore air resistance.

Water ballon and the boy start falling at the same instant.In a time t, they each fall the same vertical distance y=1/2 at2

Atg

Maximum height H

H 2 2sin2

o ovHgθ

=g

Th f h j il l i i i θ0 0

0 00 0

The y-component of the projectile velocity is: sin

sinAt point A: 0 sin

y

y

v v gt

vv v gt tg

θ

θθ

= −

= → − → =

( ) ( )22

0 0 0 00 0 0 0

sin sin( ) sin sin2 2

g

v vgt gH y t v t vg gθ θθ θ

⎛ ⎞= = − = − →⎜ ⎟

⎝ ⎠2 2sin

2o ovH

gθ

=

Atg

2 2sin2

o ovHgθ

=H

g

g

( )2 2

We can calculate the maximum height using the third equation of kinematics

for motion along the y-axis: 2v v a y y− = −( )

2

for motion along the y axis: 2

In our problem: 0 , , sin , 0 , and y yo o

o yo o o y

v v a y y

y y H v v v a gθ= = = = = − →2 2 2sinyov v θ2 2yov gH H− = − →

sin2 2

yo o ovg g

θ= =

( )

( )

x 0 0

2

0 0 0 0

v cos (eqs.1) cos (eqs.2)

sin (eqs 3) sin (eqs 4)

o ov x v t

gtv v gt y v t

θ θ

θ θ

= =

= − = − π/2

3π/2 ϕ

sinϕ

O( )0 0 0 0sin (eqs.3) sin (eqs.4)2

The distance OA is defined as the horizantal range Horizontal RangAt point A

e:

yv v gt y v t

R

θ θ

we have: 0 From equation 4 we have:y =

π/2

At point A

( )2

0 0 0 0

we have: 0 From equation 4 we have:

sin 0 sin 0 This equation has two solutions:2 2

ygt gtv t t vθ θ⎛ ⎞− = → − =⎜ ⎟

⎝ ⎠Solution 1. 0 This solution correspond to point O and is of no interest

Solution 2.

t =

0 0 sin 0 This solution correspond to point A2gtv θ − =

0 0

22 sinFrom solution 2 we get: If we substitute in eqs.2 we get:vt t

gθ

=

2 22

O A t

2 2

o

2 sin cos sin 2

has its maximum value when 45

o oo o o

v vRg g

R

θ θ θ

θ

= =

= °

Ro

2

maxovRg

=2sin cos sin 2A A A=

Example : Level horizontal range.

(a) Derive a formula for the horizontal range R of aprojectile in terms of its initial speed v0 and angle θ0. Thehorizontal range is defined as the horizontal distance the

l l b f l h hprojectile travels before returning to its original height(which is typically the ground); that is, y(final) = y0. (b)Suppose one of Napoleon’s cannons had a muzzle speedSuppose one of Napoleon s cannons had a muzzle speed,v0, of 60.0 m/s. At what angle should it have been aimed(ignore air resistance) to strike a target 320 m away?( g ) g y

Projectile motion is parabolic:

T ki h i f d f i f i dTaking the equations for x and y as a function of time, andcombining them to eliminate t, we find y as a function of x:

This is the equation for a parabola.

Examples of projectilemotion Notice the effectsmotion. Notice the effectsof air resistance.