kinematics - just physics · 2020. 12. 16. · kinematics 8 motion in 1-dimension (straight line...

JEE | NEETKINEMATICS

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KINEMATICS

DISTANCE :

Total path length.

NOTE

Its a scalar quantity so its value is always positive.

DISPLACEMENT :“Shortest distance between two points”

NOTE

Its a vector quantity so can be negative and cannot be positive.

Example-1

2R

RA B

Circular pathDistance = RDisplacement = 2R

SPEED (v) :

Distancev

Time

Average speed : av

Total Distancev v

Total Time

s1

t1

s2 s3 sn

t2 t3 tn

1 2 nav

1 2 n

s s ................ sv

t t ............. t

Case-1 : s1

t1

s2 s3 sn

t2 t3 tn

1 2 n 1 1 2 2 n nav

1 2 n 1 2 n

s s ................ s v t v t ............ v tv

t t ............. t t t t ............ t

If t1 = t2 = .................. = tn = t, then

1 2 nav

t v v .......... vv

nt

1 2 nav

v v ........... vv

n

if n = 2, 1 2av

v vv

2


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Case-2 : s1

v1

s2 sn

v2 vn

1 2 nav

1 2 n

s s ............ sv

t t ............ t

s = vt st

v

1 2 nav

1 2 n

1 2 n

s s ........... sv

s s s..........

v v v

If s1 = s2 = ............ = sn = s, then

av

1 2 n

nsv

1 1 1s .........

v v v

,av

1 2 n

nv

1 1 1...........

v v v

If n = 2, av

1 2

2v

1 1v v

, 1 2av

1 2

2v vv

v v

NOTE

For small time, we can also define instantaneous speed ds

vdt

.

Slope of distance time curve will give us the magnitude of speed.

VELOCITY (v) :

Velocity is a vector quantity, so it can be zero, negative or positive. It can be defined in two way.

1. Average Velocity a vv o r < v > : av

Total Displacementv

Total Time

[Cases can be concluded like cases of average speed]

2. Instantaneous Velocity i nv : in

dsv

dt

slope at displacement time curve

P

dt

ds

t

s

v =P

dsdt

Consider Few More Cases for Better Understanding

Case-1 : When displacement = constantv = 0

t

s

s = constant

dt0 v

dt


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Case-2 : s = linear

t

s

s = ktwhere k = constant

dsv k

dt constant

t

s

s = k t

v =dsdt

= k t

s

Case-3 : When s t² s = k t² where k = constant

t

s

s = k t

v = dsdt

= dsdt

(t²)

= 2tt

s

ACCELERATION :

“It is a vector quantity” and responsible for change in velocity (change in direction of velocity or change inmagnitude of velocity or both).

d va

dt

= slope at v t

curve

Few Examples for Better Understanding

Case-1 : If v

= constant d va 0

dt

t

v

t

a

0

C

0

Case-2 : If v

= linear, v = k t d va k

dt

constant

t

v

t

a

0

C

0

kslope = k


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Case-3 : If 2 dvv kt a 2kt

dt linear

t

v

t

a

0

v = kt²

slope = 2kt

CONCEPT BUILDERS :

Case-1 : If s = 0, v = 0, a = 0

t

s

0t t

v a

0 0

Case-2 : If s = c, ds

v 0dt

, a = 0

t

s

0t t

v a

0 0

c

Case-3 : If s = kt, ds

v kdt

constant, dv

a 0dt

t

s

0t t

v a

0 0

slope=k

k

Case-4 : If s = kt², ds

v 2ktdt

, dv

a 2kdt

constant

t

s

0t t

v a

0 0

slope=2k2k

Similarly

If s = ktn ns f t

Then n 1dsv nkt

dt n 1v f t

and n 2dva n n 1 kt

dt n 2a f t


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REVERSE CONCEPTS :

Case-1 : Relation between Displacement Velocity

dsv ds vdt

dt

2 2

1 1

s t

s t

ds vdt

2

2

1

1

ts

st

s vdt

2

1

t

2 1

t

s s vdt

2

1

t

t

s vdt

Change in position = Displacement = vdt= Area under v-t curve

t

v

t1 t2

Case-2 : Relation between Velocity and Acceleration

dva

dt dv = adt

2 2

1 1

v t

v t

dv adt

t

a

t1 t2

2

2

1

1

tv

vt

v adt

2

1

t

2 1

t

v v adt Change in velocity = v = Area under a–t curve

UNIFORMLY ACCELERATED MOTION :(Newton’s Equations of Motion)

As per the previous conceptsIf a = constantThen v = linear (v = f(t))and s = quadratic (s = f(t²))

Consider a case :Let acceleation = aat t = 0, v = u


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and at t = t, v = u and s = displacement

Case-1 : Relation between v, t & a (Newton’s 1st Equation of motion)

dva

dt dv = adt

v t

u 0

dv a dt a = constant

[v – u] = t

0a dt

v – u = a(t – 0) v = u + at ..........(1)

Slope of the curve

dv v u

adt t

t

v

t

u

v

v – u = at v = u + at

Case-2 : Relation between s, a and t (Newton’s 2nd equation of motion)

dsv

dt ds = v dt and v = u + at

ds = (u + at)dt u & a both are constant ds = udt + atdt

s

0

ds u dt a tdt

21

s ut at2

t

s

0t t

v a

0 0

a=

Case-3 : Relation between u, v and s (Newton’s 3rd equation of motion)

dva

dt

dva v

ds

v dv = a ds

v s

u 0

vdv a ds

v2

s

0u

va s

2

2 21v u a s 0

2

v² – u² = 2asv² = u² + 2as


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Example-1A body starts with speed ‘u’ under acceleration of ‘t’ where is a constant and ‘t’ is time. Its speed aftertime ‘t’ is

[a] u + t² [b] u + t² [c] u + 2t

2

[d] u + 2t²

Solution :Since ‘a’ is not uniform, so we cannot use the Newton’s equation of motion

dva

dt

dvt

dt dv tdt

v t

u 0

dv tdt

t2

v

u0

tv

2

2t

v – u2

2t

v u2

Ans. (c).


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MOTION IN 1-DIMENSION(Straight Line Motion)

“If a body is moving in a straight line and its direction is not changing, then its displacement will be equal todistance”.

DISPLACEMENT IN nth SECONDUNDER UNIFORM ACCELERATION

The above concept can be understood in following way (Straight line distance = displacement).

s1 s2

s3

s4

sn–1

sn

Let s1 = distance travelled in 1s

s2 = distance travelled in 2s

s3 = distance travelled in 3s

sn–1 = distance travelled in (n–1)s

an = distance travelled in ns

Now let th1s = distance travelled in 1st s

nd2s = distance travelled in 2nd s

rd3s = distance travelled in 3rd s

thns = distance travelled in nth s


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so, th2

1 01

1s s s ut at 0

2

nd2 2

2 12

1 1s s s u(2) a(2) u(1) a(1)

2 2

rd2 2

3 23

1 1s s s u(3) a(3) u(2) a(2)

2 2

th2 2

n n 1n

1 1s s s u(n) a(n) u(n 1) a(n 1)

2 2

th2 2

n

1 1 1s un an un u an a an

2 2 2

2 21 1 1un an un u an a an

2 2 2

1

u an a2

thn

1s u a 2n 1

2

Case-1 : When u = 0, a = constant

st1

1 1s a 2 1 1 a 1

2 2

nd2

1 1s a 2 2 1 a 3

2 2

rd3

1 1s a 2 3 1 a 5

2 2

thn

1 1s a 2n 1 a 2n 1

2 2

st nd rd1 2 3s : s : s : ........................ 1 : 3 : 5 : 7 : .................. : 2n 1

Case-2 : If u = 0, a = constant

21

1s 0 a 1

2

22

1s 0 a 2

2 s1 : s2 : s3 : ..................... = 1² : 2² : 3² : ................n²

23

1s 0 a 3

2

2n

1s 0 a n

2


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SIGN CONVENTION :

A surface or starting point can be taken as reference surface / line.

Let this surface as reference

v

v1

a

h2

h1

u

u & h1 = positive

a, v1, v & h2 = negative.

MOTION UNDER GRAVITY :

As we all know earth attracts every particle/body towards its centre. The force of attraction due to earth iscalled Gravity and acceleration produced due to it is called acceleration due to gravity. It is always towardsthe centre of gravity i.e. downwards thats why we always consider it negative.

Since a = –g = constant so we can use all 3 Newton’s equations :

Case-1 : A block / ball is dropped from a building

(a) Final velocity (v) :

u² = u² + 2as

v² = 0 + 2(–g) (–h)h

v=?

u = 0 a = –g

= 2gh

v 2gh

|v| 2gh

(b) Time taken to reach to bottom (t)

21

s ut at2

2h

tg

21h 0 gt

2

21h gt

2

Case-2 : A block / ball is thrown vertically downward

(a) Final velocity (v) : v² = u² + 2as

h

v=?

u g

v² = (–u²) – 2g(–h)v² = u² + 2gh

2v u 2gh

2|v| u 2gh



(b) Time taken to reach at bottom (t)

21s ut at

2

2u u 2ght

g

21h ut gt

2

22h 2ut gt 2gt 2ut 2h 0 .............(1)

22u 4u 4g 2ht

2g

22 u u 2gh

2g

the value inside root is greater than u and time can not be negative.

so,2u u 2gh

tg

Case-3 : A block / ball is thrown vertically up with speed ‘u’ and land at the point of projection.

(a) Time of Ascent (time taken to rise)

v = 0 at highest point

so, v = u + atu = hmax

u

v=0

0 = u – gt

u

tg

(b) Time of flight (T)It landed again at the point of intersection, hence s = 0

21

s ut at2

210 uT gT

2

2uT

g

10 T u gT

2

T = 2t

T = 0 and 1

u gT 02

2u

Tg

Let t ' = time of descentThen t + t ' = T

t t ' 2t

T ut ' t

2 g



Time of ascent (t) = Time of descent ( t ' )

= ug

This motion is symmetric i.e. upward motion is equivalent to downward motion.

(c) Final velocity (at the time of landing) v = u + at

2ut T

g (When it will land again on ground)

2uv u g

g

= u – 2uv = –u

|v| u initial speed = final speed

Case-4 : When block / ball is projected upward and landed ‘h’ depth below the point of projection.

(a) Time of flight (T)

21s ut at

2

21h ut gt

2

v

h

u

22h 2ut gt gt² – 2ut – 2h = 0

22u 4u 4g 2ht

2g

22u 2 u 2gh

2g

2u u 2ght

g

2u 2gh is greater than ‘u’ so negative sign cannot be considered

2u u 2ght

g

(This time of flight can be obtained by adding time of case-2 & case-3)

(b) Find speed (v)

v² = u² + 2as

v² = (+u)² – 2g (–h)

= u² + 2gh

2v u 2gh

2|v| u 2gh



SPECIAL CASE

A ball is projected upward with speed ‘u’ as shown.

uh

A A

OO

Find out time when it will be at ‘h’ height from the ground ?

21s ut at

2

21h ut gt

2

22h 2ut gt OA = time taken to reach from O to A (During rise)

gt² – 2ut + 2h = 0 (first time, at h displacement)

22u 4u 4 g 2g

t2g

OA = time taken to reach from O to A

(second time at h displacement)

22u 4u 8gh

2g

2u u 2ght

g

2u u 2ght

g

2

1

u u 2ght

g

2

2

u u 2ght

g

= tOA = tOA

SOME INTERESTING FACT

2 2

1 2

u u 2gh u u 2ght t

g g

2uT Time of flight

g

T = t1 + t2 = TOO

= OA AO' 1 2t t t t 1 AO ' 1 2t t t t

AO ' 2t t

AO ' A 'O ' 1 2t t t t 2 A 'O ' 2 2t t t t

A 'O ' 1t t

Graphically

uh

A A

OO

AA

t1

t2

t2t1

A A A

t2

A

t2

O OO O O O



MOTION IN A PLANE : MOTION IN 2-DIMENSION

“When a body is moving in 2-dimensional space i.e. may be in x–y or x–z ory–z plane then its motion is called 2-D motion”.

Newton’s all 3 equation of motion are valid in motion in plane also.

We should apply these equation in 1-D at a time. Following are the possible cases of 2-D. Which can also besaid as projectile motion.

Case-1 : Case-2 :

h

u

h

u

Case-3 : Case-4 :

u

O A

u

O

h

Case-5 :

u

h

Case-1 : A particle is thrown horizontally from some height.

h

ug

u

u

v

y

xO

R

(a) vx = u (remain constant, because ax = 0)

(b) vy = ? v² = u² + 2as2 2y y y yv u 2a s (y-direction motion)



v² = 0 – 2g(–h)v² = 2gh

v 2gh |v| 2gh

(c) Time of flight (T)

21

s ut at2

(y-direction motion)

21h 0 gT

2

2h = gT²

2hT

g

(d) Final speed / velocity^ ^

fv u i v j

2 2f|v | u v

2f|v | u 2gh

(e) Range (R)Horizontal distance

2

x x x

1s u t a t

2 (x-direction motion)

R = ut + 02h

R ug

( ax = 0)

Case-2 : A particle is thrown at some angle down from height

h

uu =usiny

u =ucosx

ux

ux

g

R

(a) vx = ux = u cos (ax = 0)

(b) vy = ? v² = u² + 2as2 2y y y yv u 2a s (y-direction motion)

22v u sin 2g h 2 2v u sin 2gh

2 2|v| u sin 2gh

(c) Time of flight (T)

2y y y

1s u t a t

2 (in y-direction motion)



21h usin T gT

2

22h 2usin T gT 2gT 2usin T 2h 0

2 22usin 4u sin 4g 2hT

2g

2 22usin 2 u sin 2gh

2g

22 2y yu u 2ghusin u sin 2gh

Tg g

(considered positive sign because 2y yu 2gh u )

(d) Final speed / velocity^ ^

f x yv v i v j

2x yu i u 2gh j

2 2u cos i u sin 2gh j

22 2 2 2

fv v u cos u sin 2gh

2 2 2 2u cos u sin 2gh

2 2 2u cos sin 2gh

2u 2gh

(e) Range (R) : Horizontal distance

2

x x x

1s u t a t

2 (x-direction motion)

R = uxT + 0 (ax = 0)

2 2usin u sin 2gh

u cosg

Case-3 : Simple Projectile (Ground to ground projectile)

y

0 x

y

0x

uxux=ucos

Huxu

RA

vy

ux

ux

g

uy=usin



We can start this case by collecting basic information about it

Given, initial velocity^ ^ ^ ^

i x yv u i u j u cos i u sin j

Acceleration^ ^ ^ ^

x yA a i a j 0 i g j

Velocity after time ‘t’^ ^

t x yv v i v j

^ ^

x x y ya t i u a t j u

^ ^

x y yu i u a t j

^ ^

tv ucos i usin gt j

Now we can consider the motion of projectile in 2 seperate direction i.e. in x-direction as well as iny-direction.

y-Direction Motion :(a) Time of Ascent (t) : Time to rise from bottom to highest point.

vy = uy + ayt

y0 u gt

yu usint

g g

(b) Time of flight (T) : Time taken by the projectile to cover complete path (from O to A)

2

y y y

1s u t a t

2

2y

1O u T gT

2

y

1O T u gT

2

T = 0 & y

1u gT 0

2 Time of flight = 2 (Time of Ascent)

y2u 2usinT 2t

g g

(c) Time of Descent (t’) : time taken by the particle to cover from top to bottom. time of Ascent (t) + time of descent (t’) = Time of flight (T)

t + t’ = Tt + t’ = 2t

yuT usin

t ' t2 g g

(d) Maximum Height (H) : At highest point verticalSpeed = 0

2 2y y y yv u 2a s

2y0 v 2g(H)



2 2 2yu u sin

H2g 2g

x-Direction Motion : (ax = 0)

Range (R) : Horizontal displacement covered by the body during complete motion.

2

x x x

1s u t a t

2 [ax = 0]

R = uxT

yx

2uR u

g

2x y2u u u sin2

Rg g

x y2u uR

g

2u cos usinR

g

2u 2sin cosR

g

2u sin2R

g

Few Additional Information

Final Speed : ^ ^

f x yv v i v j

vx = ux = u cos [ax = 0]& vy = uy + ayt

= uy – gt = u sin – gt

= 2usin

usin gg

2usint T

g

u sin 2u sin

y yv usin u

Final Speed : ^ ^

f x yv u i u j

^ ^

u cos i u sin j

2 2 2 2f|v | u cos u sin

2 2 2u cos sin

|v | u

Final speed = initial speed = u Motion of the above projectile is symmetrical



Example-1Find out the time taken by the projectile when it becomes perpendicular to its initial direction.

Solution :

y

0 x

u P

v

initial direction

Let after time ‘t’ from the projection its velocity becomes perpendicular to its initial direction. (Let at P)

Initial velocity^ ^

iv ucos i u sin j

Velocity after time (t) ^ ^

tv ucos i usin gt j

Now since i tv v then

i tv .v 0

^ ^ ^ ^

ucos i usin j . u cos i u sin gt j 0

2 2u cos usin usin gt 0 2 2 2 2u cos u sin usin gt 0

2 2 2u cos sin usin gt

u2 = u sin gt

ut

g sin

EQUATION OF TRAJECTORY OF PROJECTILE :

vx = u cos [ax = 0]

x

dxv ucos

dt

y

0 x

uP(x,y)

R

dx u cos dt x t

0 0

dx ucos dt x u cos t

xt

ucos

.........(1)

y yv u gt



dyusin gt

dt

dy usin dt gtdt

dy usin dt g tdt 2t

y u sin t g2

.........(2)

From (1) & (2)2

x 1 xy usin g

ucos 2 u cos

2

2 2

gxy x tan

2u cos

With the help of this equation we can find the corresponding value of x- when y- is given or vice-versa.

USES OF EQUATION OF TRAJECTORY OF PROJECTILE

2

2 2

gxy x tan

2u cos

.........(1)

at A, x = R, y = 0

2 2

gR0 R tan

2u cos

y

O A

P(x,y)

(R,0)(0,0)

H2 2

sin gR0 R

cos 2u cos

2 2

sin gRR 0 & 0

cos 2u cos

2

gRsin

2u cos

22u sin cos

Rg

(sin 2 = 2sin cos )

2u sin2R

g

Alternatively from eqn. (1)

2 2

gxy x tan 1

2u cos tan

2

xx tan 1

2u sin cosg

xy x tan 1

R



NOTE

Whenever you have equation of trajectory ofprojectile you can put y = 0, then x = R

You can also take first quantity common, then equation will be like x

y x tan 1R

and you

can easily calculate range of projectile.

Example-1

If equation of trajectory of a projectile is given as 21y x 50x

3 then find its range ?

Solution :Method-1 Method-2

21y x 50x

3 1

y x 1 50 3x3

We know y = 0, x = R 1 x

x 113

50 3

210 R 50R

3

1R

50 3 unit

1R

50 3 unit

xy x tan 1

R

CONCEPT :

Projectile is passes through a building whose angle of inclination from point of projection is and angle ofdepression from landing point is .

From equation of trajectory

xy x tan 1

R

y

0 x

P(x,y)

y

x R–x

u

R

y R xtan

x R

yR

tanx R x

..........(1)

From the diagram

y ytan , tan

x R x

After adding



y ytan tan

x R x

y R x x

x R x

yR

x R x

tan tan tan from ......(1)

Example-2A canon is fired to hit a target at certain distance. If it is fired at an angle then it misses the target by ‘a’

distance. If it is fired at angle ‘’ then it landed ‘b’ distance away from it. Find out exact angle of projection.

Solution :Let = angle of projection at which it will hit the target

2u sin2R

g

........(1)

Targeta b

R

2u sin2R R a

g

.........(2)

2u sin2R R b

g

.........(3)

Eqn. (2) × b2u bsin2

Rb abg

Eqn. (3) × a2u a sin2

Ra abg

2u

R a b bsin2 a sin2g

..........(4)

Replacing R from (1)

2 2u sin2 u

a b bsin2 a sin2g g

bsin2 asin2

sin2a b

1 bsin2 a sin 2

2 sina b

11 bsin2 a sin 2

sin2 a b



Relation between R, H & T :-2 2 2u sin2 u sin 2usin

R , H & Tg g g

Relation between R & H :-2 2

2 2 2 2

R u sin2 2g u 2sin cos 2gH g u sin g u sin

4

R tan 4Htan

..........(1)

Relation between H & T :-2 2 2

2 2 2

H u sin gT 2g 4u sin

21

H gT8

1

g8

21

4H gT2

..........(2)

From Eqn. (1) & (2)

21R tan 4H gT

2

WORKING ON RANGE OF PROJECTILE :

2u sin2

Rg

Maximum Range :

For R = Rmax

dR0

d

or

0

0

2 90sin 2 maximum 21

454

2

max

uR

g when R = max, R tan = 4H, Rmax = 4H, tan 450 = 1

Range at two complimentary angles :

= ,2u sin2

Rg

, = ,

2u sin2R

g

If + = 900 Complimentary angle

= 900 – .

2 0 2 0u sin 90 u sin 180 2R

g g

2u sin2

g

[sin(1800 – ) = sin ]



R R

R

y

x300 600 450

0

R

y

x

u

0

u

In case of complimentary angle, angle made by the projectile one with horizontal and one withvertical will remain same.

Range can also be determined with the help of equation of trajectory by following way

2

2 2

gxy x tan

2u cos

Range2

Coefficient of xR

Coefficient of x

2 2

2 2

tan sin 2u cosg cos g

2u cos

22u sin cos

g

2u sin2R

g

Example-1 :

Equation of trajectory of a projectile is given 21y 3x x

50 . Find its range ?

Solution :Method-1

We can compare the above equation with standard equation of trajectory2

2 2

gxy x tan

2u cos

and then can use 2u sin2

Rg

to get answer



given 21y 3x x

50

then 03 tan 60 .........(1)

and 2 2

g 12u cos 50

let g = 10

2u²cos² 600 = 500

2

2 1u 250

2

u² = 1000

u 10 10 m / s .........(2)

2 2u sin2 u 2sin cos

Rg g

0 01000 2 sin60 cos60 3 1

100 210 2 2

R 50 3Method-2

We can write the given equation in terms of x

y x tan 1R

and then compare it

Given 21 xy 3 x x 3x 1

50 50 3

after comaring R 50 3

Method-3

21

y 3x x50

2

Coefficient of xRange

Coefficient of (–x )

3R m

150

R 50 3 m

Case-4 : A particle is projected from ground and landed to some height.

“All initial values and acceleration

u

O

h

xu =ucosx

Range(R)

u =ucosy

yare as per the other cases”.

Vertical Motion

(a) Time of flight (T) uy = u sin

2

y y y

1s u t a t

2



2y

1h u t gt

2

2h = 2uyt – gt²

gt² – 2uyt + 2h = 0

2

y y2u 4u 4g 2ht

2g

2

y y2u 2 u 2gh

2g

2y yu u 2gh

tg

2

y yu u 2 g h

t =

g

2y y

1

u u 2ght

g

2y y

2

u u 2ght

g

=tOA

u

O

h

x

y

A

=tOB

u

O

h

x

y

A B

t1 = time taken to reach at h(vertical)displacement first time = tOA

t2 = time taken to reach at h(vertical)displacement second time = tOB

Time taken to reach from A to BtAB = t2 – t1 = t

2 2

y y y yu u 2gh u u 2gh

g g

2y

AB

2 u 2ght

g

In all three values t1, t2 and t, if we put h = 0, we will get results of case-3.



Range (R) : ax = 0 &

2x x x

1s u t a t

2 u

O

h

xR

y

B

R = uxt

R = uxt2

Where 2

y y

2

u u 2ght

g

, uy = u sin , ux = u cos .

Case-5 : A particle is projected from same height and landed to ground.

Range=R

h

u

y

x

“All basic parameters are like previous cases.

Vertical Motion

Time of Flight (T)

2

y y y

1s u t a t

2

2y

1h u t gt

2

–2h = 2uyt – gt²

gt² – 2uyt –2h = 0

2

y y2u 4u 4g 2ht

2g

2y y2u 2 u 2gh

2g

2

y yu u 2ght

g

Only positive sign can be considered, else time will be negative

2y yu u 2gh

T tg

Horizontal Motion :Range (R) ax = 0

& 2x x x

1s u t a t

2

R = uxt + 0

kinematics - just physics · 2020. 12. 16. · kinematics 8 motion in 1-dimension (straight line...

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