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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
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(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
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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
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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
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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)
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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)
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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
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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)
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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
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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
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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
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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
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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
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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 ]
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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
JEE | NEETKINEMATICS
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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
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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.
JEE | NEETKINEMATICS
27www.justphysics.in
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
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