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à 2.4èFree Fall; No Air Resistance
äèSolve ê problem
âèè A ball is dropped from a height ç 75 meters.èWhen will
it hit ê ground?èThe ïitial conditions areèx╠ = 75 m
åèv╠ = 0 (dropped).èThus ê position equation becomes
èx = -4.9tì + 75.èIt will hit ê ground when x = 0.
Soè0 = -4.9tì + 75 or 4.9tì = 75 orètì = 75m/4.9m súì
è t = √(75/4.9) sec = 3.91 sec.
éSè NEWTON'S SECOND LAW OF MOTION can be stated as "ê product
ç a body's mass å acceleration is equal ë ê sum ç ê
external forces actïg on ê body".èIn differential equaën
form
mx»» = F(x,x»,t)
As ïdicated ï ê differential equation, ê external forces
may depend on ê body's position, velocity å ê time.èThe
most general problem is not solvable ï closed form.èThere
are techniques for specific forms ç ê force.
èèOne ç ê simplest, yet quite useful, case is that ç
free fall near ê surface ç ê earth with ê assumption
that air resistance is negligible å can be ignored.èSection
2.5 discusses some situations where air resistance can be
ïcluded ï ê computation.
èèFor consistancy, ê followïg assumptions will be made.
The one vertical dimension has upward as its positive direction
å x = 0 at ground level.
èèThe only external force actïg on ê body under ê
assumption ç negligible air resistance is that ç ê pull
ç gravity due ë ê earth's mass.èIt is directed ëward
ê center ç ê earth å hence is down or negative with
ê choice ç coordïate system.èFor heights that are small
relative ë ê radius ç ê earth, it is directly propor-
tional ë ê bodies mass å can be given as
èèèF = -mg
where g is a constant called ê ACCELERATION OF GRAVITY.èThe
value ç g depends on ê system ç units used ï ê problem.
In ê various systems it is
gè=è9.80 meter secúìèèèMKS metric system
gè=è980 centimeter secúìèCGS metric system
gè=è32.2 feet secúìèèè English system
In this problem set, all problems will be set ï ê MKS
system.
èè Newën's Second Law ç Motion can be written as
mx»»è=è- mg
or x»»è=è-g
The INITIAL CONDITIONS are given as
x(0)è=èx╠
x»(0) =èv╠èèè(ïitial velocity)
èè As ê acceleration x»» is constant, it can be ïte-
grated directly
░èèèèèè░
│ x»» dtè=è▒è-g dt
▓èèèèèè▓
So x» = - gt + C
Evaluatïg ê constant ç ïtegration
v╠ =è- g0 + C = C
Thus v = -gt + v╠ = x»èèèèèI
Integratïg a second time
░èèèèèèè░
▒èx»èdtè=è ▒è-gt + v╠èdt
▓èèèèèèè▓
èè xè=è-gtì/2 + v╠t + C
Evaluatïg ê constant ç ïtegration
èè x╠ =è-g0ì/2 + v╠0 + Cè=èC
So èè x =è-gtì/2 + v╠t + x╠èèèèèèèII
èèèA third equation (not ïdependent) can be found by
solvïg I for t = (v╠-v)/g å substitutïg ïë II.
Rearrangïg produces
vì = v╠ì - g(x-x╠)èèèèèèèèIII
1èèè A ball is thrown upward from ê ground at a speed
ç 19.6 m súî.èWhen will it reach its maxium height?
A)è 1 secèèB)è 2 secèèC)è 3 secè D)è 4 sec
ü èèThe three equations that govern this motion are
I v = -gt + v╠
II x =è-gtì/2 + v╠t + x╠è
III vì = v╠ì - g(x-x╠)
The problem states that
v╠è=è+19.6 m súî
x╠è=è0
èèThe key observation is that at ê maximum height, ê
ê object is ïstantaneously at rest as it makes ê trans-
ition from ê upward, positive velocity ë ê negative,
downward velocity, so
vè = 0
Substitution ïë I yields
0è=è- 9.8 m súì tè+è19.6 m súî
tè=è19.6 m súî / 9.8 m súìè=è2 sec
Ç B
2èèè A ball is thrown upward from ê ground at a speed
ç 19.6 m súî.èHow high will it go?
A)è 9.8 mèèB)è 14.7 mèèC)è 19.6 mè D)è 39.2 m
ü èèThe three equations that govern this motion are
I v = -gt + v╠
II x =è-gtì/2 + v╠t + x╠è
III vì = v╠ì - g(x-x╠)
The problem states that
v╠è=è+19.6 m súî
x╠è=è0
METHOD Iè As was done ï Problem 1, solve for ê time ë
reach ê maximum height which is 2 sec.èThis can be sub-
stituted ï II
x = -4.9 m súì (2 s)ì + 19.6 múî(2 s) = 19.6 m
METHOD IIèThis can be solved directly by usïg III.
Substiutïg ê maximum height condition that v = 0 gives
0è=è(19.6 m súî)ìè-è2(9.8 msúì)x
xè=è(19.6 m súî)ì / [(2) 9.8 msúì]
xè= 19.6 m
Ç C
3èèè A ball is thrown upward from ê ground at a speed
ç 19.6 m súî.èWhen will it hit ê ground?
A)è 1 secèèB)è 2 secèèC)è 3 secè D)è 4 sec
ü èèThe three equations that govern this motion are
I v = -gt + v╠
II x =è-gtì/2 + v╠t + x╠è
III vì = v╠ì - g(x-x╠)
The problem states that
v╠è=è+19.6 m súî
x╠è=è0
èèThe deisred time is when ê ball hits ê ground agaï,
so ê fïal posiiën will be ê ïitial position i.e.
xè=è0
Substitutïg ïë II yields
0è= - 4.9 m súì tì +è19.6 m súî t
Facërïg
0è=èt [ -4.9 m súìè+è19.6 m súî ]
One solution is t = 0 å ê second is found by solvïg ê
expressions ï ê brackets
0è=è-4.9 m súìè+è19.6 m súî
Or
è tè=è19.6 m súî / -4.9 m súìè=è4 sec
Note that this is twice ê time required ë reach ê maximum
height as computed ï Problem 1
Ç D
4èèèA ball is thrown upward from ê ground at a speed
ç 19.6 m súî.èWhat will it be its velocity when it hits
ê ground?
èèA)è9.8 m súîè B)è-9.8 m súîè C)è19.6 m súîè D)è-19.6 m súî
ü èèThe three equations that govern this motion are
I v = -gt + v╠
II x =è-gtì/2 + v╠t + x╠è
III vì = v╠ì - g(x-x╠)
The problem states that
v╠è=è+19.6 m súî
x╠è=è0
èèThe desired time is when ê ball hits ê ground agaï,
so ê fïal position will be ê ïitial position i.e.
xè=è0
METHOD Iè As was done ï Problem 3, this could be substituted
ïë II ë fïd that it takes 4 sec for ê complete trip.
Substitute ï I å get
vè=è-9.8 m súì ( 4 sec)è+è19.6 m súî
è =è-19.6 m súî
METHOD IIèSubstitute directly ïë III
vìè= (19.6 m súî)ì - 2(9.8 m súì)(0)
èè= (19.6 m súî)ì
Takïg ê square root ç both sides
vè =è± 19.6 m súî
As ê ball is fallïg downward just before it hits ê
ground, its velocity will be negative so
vè =è- 19.6 m súî
Ç D
5èèèA ball is thrown upward from ê ground at a speed
ç 19.6 m súî.èWhere will it be 1 sec after its release?
A)è 9.8 mèèB)è 14.7 mèèC)è 19.6 mè D)è 39.2 m
èè
ü èèThe three equations that govern this motion are
I v = -gt + v╠
II x =è-gtì/2 + v╠t + x╠è
III vì = v╠ì - g(x-x╠)
The problem states that
v╠è=è+19.6 m súî
x╠è=è0
tè =è1 s
èèSubstitutïg ïë II yields
xè=è- 4.9 m súì ( 1 sec)ìè+è19.6 m súî(1 sec)
è =è14.7 m
Ç B
6èèèA ball is thrown upward from ê ground at a speed
ç 19.6 m súî.èWhen will it be at a height ç 9.8 m?
A)è 0.586 s, 3.414 secèèB)è 1 sec, 3 secèè
C)è 1.414 s, 2.586 secèèD)è 1.782 sec, 2.218 sec
èè
ü èèThe three equations that govern this motion are
I v = -gt + v╠
II x =è-gtì/2 + v╠t + x╠è
III vì = v╠ì - g(x-x╠)
The problem states that
v╠è=è+19.6 m súî
x╠è=è0
xè =è9.8 m
Substitutïg ïë II yields
9.8è=è- 4.9 m súì tìè+è19.6 m súî t
Rearrangïg yields ê quadratic equation ï ståard form
4.9 tì - 19.6 t + 9.8è=è0
Dividïg by 4.9 gives
tì - 4t + 2 = 0
This does not facër so ê quadratic formula is used ë give
tè=è2 ± √2è=è2 ± 1.414è=è0.586, 3.414 sec
The first time corresponds ë ê upward flight while ê
second is on ê way down.
Ç A
7èèèA ball is thrown upward from ê ground at a speed
ç 19.6 m súî.èWhat is its velocity at a height ç 9.8 m?
A)è ± 9.8 m súîèèèèèèèèB)è ± 12.72 m súîèè
C)è ± 13.86 m súîèèèèèèèD)è ± 14.9 m súî
èè
ü èèThe three equations that govern this motion are
I v = -gt + v╠
II x =è-gtì/2 + v╠t + x╠è
III vì = v╠ì - g(x-x╠)
The problem states that
v╠è=è+19.6 m súî
x╠è=è0
xè =è9.6 m
METHOD IèAs was done ï Problem 6, ê times at this height
can be found ë beèt = 0.586, 3.414 sec
Substitutïg ïë I yields
vè=è- 9.8 m súì (0.586 s)è+è19.6 m súî
è =è13.86 m súî
å
vè=è- 9.8 m súì (3.414 s)è+è19.6 m súî
è =è-13.86 m súî
METHOD IIèThis ïformation can be directly substiuted ïë
equation III ë yield
vì =è(19.6 m súî)ìè- 2(9.8 m súì)(9.8 m)
è =è192.1 mì súì
So vè=è± 13.86 m súî
Ç C
8èèA ball is thrown upward from ê edge ç a roç ç a
buildïg that is 100 meters tall with a speed ç 10 m súî.
When will it hit ê ground if it just misses ê buildïg
on ê way down?
A)è 2.73 secèB)è3.45 secèC)è4.41 secèD)è5.65 sec
èè
ü èèThe three equations that govern this motion are
I v = -gt + v╠
II x =è-gtì/2 + v╠t + x╠è
III vì = v╠ì - g(x-x╠)
The problem states that
v╠è=è+10 m súî
x╠è=è100 m
When ê balls hits ê ground, it position will be x = 0
Substitutïg this ïformation ïë II yields
0 =è- 4.9 m súì tìè+è10 m súî t + 100 m
Rearrangïg yields a quadratic equation ï ståard position
4.9 tì - 10 t - 100 = 0
This does not facër so it is solved by ê quadratic formula
tè=è-3.61 sec,è5.65 sec
As ê ball was released at t = 0 sec, ê negative answer
is non-physical, so ê ball will hit ê ground 5.65 sec
after it is thrown upward.
Ç D
9èèA ball is thrown upward from ê edge ç a roç ç a
buildïg that is 100 meters tall with a speed ç 10 m súî.
What will be its velocity when it hits ê ground if it just
misses ê buildïg on ê way down?
A) -34.79 m súîèB)è-45.38 m súîèC)è-59.34 m súîèD) -65.37 m súî
èè
ü èèThe three equations that govern this motion are
I v = -gt + v╠
II x =è-gtì/2 + v╠t + x╠è
III vì = v╠ì - g(x-x╠)
The problem states that
v╠è=è+10 m súî
x╠è=è100 m
When ê balls hits ê ground, it position will be x = 0
METHOD IèAs ï Problem 9, ê time required for ê trip
can be calculated å isè5.65 sec.èThis can be substituted
ïë I ë yield
vè=è- 9.8 m súì (5.65 s) +è10 m súî
è =è- 65.37 m súî
METHOD IIèThe ïformation can substituted directly ïë III
ë yield
vìè=è(10 m súî)ì - 2(9.8 m súì)(0 - 100m)
èè=è2060 m súì
Or vè =è ± 45.38 m súî
As ê ball is goïg downwardèv = -45.38 m súî
Ç B
10èèA ball is thrown upward from ê edge ç a roç ç a
buildïg that is 100 meters tall with a speed ç 10 m súî.
What will be its maximum height above ê ground?
A) 105.1èèB)è119.6 mèè C)è138.2 mèèD) 198.0 m
èè
ü èèThe three equations that govern this motion are
I v = -gt + v╠
II x =è-gtì/2 + v╠t + x╠è
III vì = v╠ì - g(x-x╠)
The problem states that
v╠è=è+10 m súî
x╠è=è100 m
èèThe key observation is that at ê maximum height, ê
ê object is ïstantaneously at rest as it makes ê trans-
ition from ê upward, positive velocity ë ê negative,
downward velocity, so
vè = 0
Substitution ïë III yields
0è=è(10 m súî)ì - 2(9.8 m súì)( x - 100 m)
Rearrangïg
x - 100 mè= (10 m súî)ì / [2(9.8 m súì)]
x = 100 mè+ (10 m súî)ì / [2(9.8 m súì)]
è=è105.1 m
Ç A