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Projectile Motion

DISCLAIMER
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to Sorting Hat Technologies Private Limited (“Company”)
...
The author of the content shall be solely
responsible towards, without limitation, any claims, liabilities, damages or suits which
may arise with respect to the same
...
Projectile motion is an
example of 2-D motion
...
Oblique or angular projectile motion
...
Horizontal projectile motion
...


Definitions
When a body is thrown at an
angle from horizontal then
motion under gravity is known
as oblique projectile motion

Concept Reminder

The shape of the trajectory
of motion of an object is not
determined by acceleration alone
but also on initial conditions (like
initial velocity)
...


1
...


Projectile Motion

Projectile Motion

2
...

For x-axis:
1
x  ux t  a x t 2
2

KEY POINTS

 ax  0

x  ucos t  t 

x
u cos 


...
(ii)

Rack your Brain

From equation (i) and (ii)
y  u sin  

x
1  x 
 g

u cos  2  ucos  

y  x tan  

gx2
2u2 cos2 

2

The equation of projectile is
5x2

...

Let tan   A and

g
2

2u cos2 

B

So, y  Ax  Bx2

3
...



y  x tan  

gx2
2u2 cos2 


x
y  x tan   1  
 R

Concept Reminder

Vertical component of velocity
is zero when particle moves
horizontally i
...
, at the highest
point of trajectory
...
O
...
O
...
 ucos ˆi  u sin ˆj
(ii)

Rack your Brain
Two seconds after projection
a projectile is travelling in a
direction inclined at 30° to the
horizontal after one more sec,
it is travelling horizontally, what
are the magnitude and direction
of its velocity ?

At T
...
P
...
O
...
 u cos ˆi  0ˆj

4
...
O
...
(Point of Impact)

Concept Reminder:
At highest point of projectile
av


vP
...
I
...


Time of Flight:
The time taken by projectile between point of
projection and point of impact is known as time
of flight
...

Find its velocity of point of
impact
...


5
...


If time taken is T during motion between points
A to B


1
So, Sy  uy T  a y T2
2


g
0  uy T  T2
2
T

2uy
g

2u sin 
T
g

Definitions
During
projectile
motion
maximum height attained by
object on vertical axis is known
as maximum height
...


Rack your Brain
For a projectile, what is the ratio
of maximum height reached to
the square of flight time?
At maximum height
vy  0
so, v 2y  u2y  2a y Sy
0  u2y  2gH
H
H

u2y
2g
u2 sin2 
2g

at   90
H becomes maximum, so greatest height

6
...


Concept Reminder

For oblique projectile
* T
* H

2uy
g

u2y
2g

* R  U T 

2u  uy
g

Time taken to reach at point from point A is T
(time of flight)
...
Find the angle of
projection
...


Projectile Motion

R  ux 

Keywords

R

u2 sin 2
g

Concept Reminder

at  = 45° range becomes maximum so,
Rmax 

u2 sin 2  45 u2
u2

 Rmax 
g
g
g

Value of height when range is maximum,
h

It was Gallileo who first stated the
principle of physical independence
of the horizontal and vertical
motion of a projectile in his book
“Dialogue on the great world
system” in 1632

u2 sin2 45 u2

2g
4g

Relation between R and H:
R

u2 sin 2
g


...
(ii)

From (i) and (ii)
u2 sin 2
g

R

H u2 sin2 
2g

Concept Reminder:
Relation between R and H:
R tan   4H

R
4

H tan 

Projectile Motion

R tan   4H

8
...


A1

According to question
R  nH
u2 sin 2 n  u2 sin2 

g
2g
tan  

4
n

4
  tan1   (Angle of projection)
n
if R = 4H
4
Then angle of projection is   tan1    1
n
tan   1
  45
If R = 2H
4
tan  
2

 

  tan1 2
2
5

, sin  

1
5

9
...
The
equation of trajectory is y  3x 

gx 2

...
Angle of projection
2
...
Time of flight
4
...
Range
6
...


A2

Equation of trajectory
y  x tan  
y  3x 
1
...


u2 cos2   1
u2  cos2 60  1
1
u2   1
4
u  2m s

Projectile Motion

3
...
2 3 sec
...


H

u2 sin2 
43

 0
...



x 


gx
2 3

y  3x 
 3 1

2
g 




2

R

2 3
 0
...


6
...
Find (i) Angle of projection
(iii) Range
(v) Maximum height

A3

(ii) Time of flight
(iv) Initial speed

Initial velocity

u  uxˆi  uyˆj  6iˆ  8ˆj
u cos   6

1
...


Time of flight
T

3
...
6 sec
...
6 m
10

4
...


Hmax 

u2y
2g



82
 3
...


Projectile Motion

Q3

3
 3m s
2

Q4

A body of mass m is thrown at angle  with horizontal from ground with
initial velocity v
...
Velocity is half of initial velocity
...
Velocity is

A4

3
times of initial velocity
2

Velocity at highest point = u cos 
1
...


u cos  

Þ

  30

u
2
3u
2

Q5

A body is thrown at angle  from horizontal with initial velocity u
...


1
...

2
...


Let at point P, velocity becomes perpendicular to initial velocity
 
So, u  vp  0

u cos ˆi  u sin ˆj  ucos ˆi  u sin   gt  ˆj  0
u2 cos2   u2 sin2   u sin gt  0





Projectile Motion

u2 cos2   sin2   u sin gt  0

2
...


Important Concept
Comparisons of Two Projectiles of Equal Range:
When two projectile are thrown with equal speeds
at angle  and 90   then their ranges are equal
but maximum height attained are di erent and
time of flight are also di erent
...


q

90 – 

R1 and R2

30°

60°

R 1 = R2

60°

30°

R 1 = R2

16°

74°

R 1 = R2

74°

16°

R 1 = R2

Definitions
According to Galileo, in his book
“Two New Sciences” stated
that for elevations by which
exceed or fall short of 45° by
equal amounts, the ranges are
equal
...


H1
H2



H1
H2

sin2 
cos2 

 tan2 

Concept Reminder:
At angle  and 90 – :

 tan2 

H1  H2 

4

 R1  R2
2



2

u sin  cos 
4g 2
2

 2u2 sin   cos  
1 1 R2

   


g
4 4 16


 H1H2 

R2
16

u2 sin2  u2 cos2  u2


2g
2g
2g

H1  H2 

u2
2g

(Maximum height when thrown vertically upward
from ground)
Time of Flight of Projectiles at (q) and (90–):

T2

Projectile Motion







2u sin 90  
2u sin 
2ucos 
, T2 

g
g
g


T1
T2

sin 
 tan 
cos 
 tan 

R2
16

 H1  H2 

u2
2g

T1
 tan 
T2

 T1  T2  R

H1  H2 

T1

 H1H2 



R  4 H1H2

T1 

H1
 tan2 
H2

Rack your Brain
Two stones are projected with
the same magnitude of velocity,
but making di erent angles
with horizontal
...


T1T2 

4u2 sin  cos 
g

2



2R
g

T1  T2  R

Concept Reminder

AIl projectile will go upto the
maximum horizontal distance,
if it is projected at 45° with the
horizontal
...

u2 sin 2  45 u2

g
g

 Rmax 

u2
g

15
...


A6

Range is maximum at   45
H

R
u2 sin2  u2 sin2 45 u2


 max
2g
2g
4g
4

Q7

A person can throw a body to attains maximum range
...
Then what is relation
between maximum height and maximum range
...
(ii)

...


Q8

Find the relation between -

(1) RA, RB and RC
(3) HA, HB and HC

A8

1
...


H

3
...


(2) TA, TB and TC
(4) uA, uB and uC

u    u   u 
y A

y B

y C

u2y
2g

2uy
g

RC  RB  RA
R

2uxuy
g

ux C  ux B  ux A

17
...
Find1
...
Initial speed

T=

2u sin 
 10
g

u sin  

10  10
 50 m s
2

 

Range R  200 

200 



2 u sin 

tan  



2

u2 sin 2 2u2 sin  cos 

g
g

 cos 

g sin 

 

2  50

2

10  200



5000 5

2000 2

 

2

50
u2 sin2 
H

 125m
2g
2  10

Projectile Motion

sin  

5
29

T

2u sin 
g

u

gT
2 sin 

u

10  10
29  10 29
25

18
...
Find, angle of projection and initial speed
...


gx 2

y  x tan  

2u2 cos2 

x
y  x tan   1  
R


4
4  4 tan   1 

18 

 14 
1  tan   
 18 
18 9

14 7
9
  tan1  
7
tan  

2
...


Change in Linear Momentum:
We know that change in momentum

 
P  m vF  vI





Now we discuss about change in momentum
between di erent points on a projectile path
...

2
...

4
...
O
...
O
...
O
...
O
...
Between P
...
P and T
...
P:

P  mu sin ˆj



Rack your Brain
A particle of mass 100 g is
fired with a velocity 20 m sec–1
making an angle of 30° with the
horizontal
...


Concept Reminder

Change in momentum of projectile
during its half journey and full
journey is
-Musin ĵ and -2 musin ĵ
respectively
...
Between P
...
P and P
...
I :

P  2mu sin ˆj
3
...
O
...
O
...



u sin 
 P  mu sin  ˆj
g

2u sin 
tT
 P  2mu sin  ˆj
g

Angular Momentum j :t top 



It is moment of linear momentum
...


Concept Reminder

If in case of projectile motion range
R is n times of maximum height H
i
...
, R = nH
Then angle of projection
4
  tan1  
n

Keywords
 Linear momentum
 Angular momentum
d
 sin 
R
d   R sin 
J  P  d   PR sin 

 
J  RP



 
J  R,
 
J  P,



 
JR  0
 
JP  0

Rack your Brain
A particle is projected from
ground with speed 80m/s at
an angle 30 with horizontal
from ground
...


21
...
Find angular momentum about point of projection
1
...
In above is question if angle of projection are (a) 45°
(b) 60°
3
...
O
...


A11

1
...


(a)

v 2 sin2 
mv 3 sin2  cos 
 mv cos  
2g
2g
J0 
J0 

(b)

3
...


Q12

Projectile of mass m is thrown with initial velocity u at an angle  with
horizontal from ground
...
O
...


A12

tan  
tan  
tan  

2H
R
2  u2 sin2   g
2g  2u2 sin  cos 
tan 
2

Kinetic energy, gravitational, potential energy and total mechanical energy
when projectile thrown from grown:-

(i)

G
...
E  mgh

(ii)

KE 

1
mu2
2

23
...
E  K
...
E

A
...
O
...

G
...
E = 0
1
2
...

Total M
...


T
...
P (H = H)
mgu2 sin2 
 E0 sin2 
2g

1
...
P
...


K
...


Total M
...
P
...
E
C
...
O
...


G
...
E  0

2
...
E 

1
mu2  E0
2

3
...
E = E0
Total M
...
at every point (P
...
P, T
...
P and P
...
I) is same
...
O
...


A13

1
mv 2  E0  4J
2

KE TOP  E0 cos2 
Projectile Motion

 4

1
 2J
2

P
...
O
...


Q14

Find angle of projection when kinetic of energy at topmost point is

3
times
4

of initial kinetic energy
...
ET
...
P  3 K
...
O
...


Projectile Motion

1
3 1
 u2 cos2    mu2
2
4 2
3
cos2  
4



Direction of velocity of projectile when thrown
by making angle q with horizontal from ground:-

v 
v 

p y

tan  

v 
v 

p x
p y

1
...
O
...
P
...
I t  T 

2u sin 
= Time of flight
g

Concept Reminder


In terms of H :tan  

v 
v 

p y



p x

h

tan   tan   tan 

u2 sin2   2gh
,
u cos 

u2 sin2 
2g

At T
...
P
v 2y  u2y  2a y sy ,

tan   0



26
...
P
...
P and T
...
P
2
...
O
...
O
...


v =

Total displacement
Total time
2

R 
2
  H
2
 
T

v =

2

v 

 2u2 sin  cos  
 u2 sin2  

 

2g
2g




u sin 
g

v 

u4 sin2  
4 cos2   sin2  
2


4g
u sin 
g



u2 sin 
2g
v 
u sin 
g

2
...


Projectile Motion

Q15

Horizontal Projectile:When a projectile thrown from certain height H
horizontal direction then


ux  u ,
ax  0


uy  0 ,
a y  g
1
...
sy  uy t  a y t 2
2

 x2
1
so, sv   g   2
2  v
t






1
...
e
...


When a projectile is thrown from height H
horizontally and dropped from same height
strike to the ground in same time
...


Concept Reminder

In Horizontal projectile
T

2H
g

R  uT  u

2H
g

 
1 
sx  ux t  a x t 2
2
R  uT  u

2H
g

3
...
A
bullet pierces A and them B
...
If the bullet is travelling
horizontally at A calculate the
velocity of bullet at A
...


Concept Reminder

The path of a projectile as seen
from another projectile is a
straight line
...


30
...
when it is at height 19
...

standing on ground a bomb will strike to the person
...


A16

1
...


T

3
...
6
 2 sec
...
8

31
...
If the steps are 0
...
Find
value of n
...


Q17

A person is throwing a ball in horizontal direction from a 18m height building
...
2m
height from ground
...
Then calculate
initial velocity of ball
...
2
H  9
...
2m
Ru

2H
g

6
...
8
9
...
2
2

6
...
4 m s
1
...


When bullets are fired with same velocity u from
stair give in di erent direction then maximum
area covered by any bullet
...


Time of flight (T): 
1 
sy  uy t  a y t 2
2
H  u sin T 

Concept Reminder

If an object is dropped from
horizontally
flying
aeroplane
the object falls in a straight line
vertically down with respect to
pilot/plane
...


1 2
gT
2

2H  2u sin T  gT2
gT2  2u sin T  2H  0

T

u sin   u2 sin2   2gH
g

Range:
 
1 
R  sx  ux t  a x t 2 , R  u cos T
2
Maximum time when strike to the ground
...


t2 

2h
g



2 Hh



g

T  t 1  t2 

2h

g



2 Hh



g



 


dT
d  2h  d  2 H  h



dh dh  g  dh 
g

dT

dh

2
g

d h d Hh 



 dh
dh 


dT

dh

2
g

 1
1 1 



2 h 2 Hh 








 

dT
0
dh

35
...





2 1
1




g  2 h 2 H  h 

0

2

2


1 2 
1
2


  
2  gh 
2 g Hh 


2
2

gh g H  h









Hh  h
2h  H
h



H
2

Collision of two projectile :Check:1
...
Step - II:
(i)

x  R 1  R2

(ii)

x  R 1  R2

(iii)

x  R 1  R2

Both called the v and V:-

UA y  UB y

Projectile Motion

v sin60  V sin 90
 3
V
v
 2 


v
2

V
3

Keywords
 Collision

36
...


uA y  uB y
10 sin 30  5 2 sin 45
10 5 2

2
2
55
R1 

2u2 sin  cos 
g

R2 

2  u2 sin  cos 
g

R1  2 
R2 

100 1
3
 
10 2 2

2  25  2
1
1


10
2
2

R1  5 3
R2 

50
5
10

R1  1
...
660
R1  R2  8
...
6
R 1  R2  x
Collision not occur
...


Projectile Motion

2
...
The observer B sitting on road
will see the ball moving in a parabolic path
...


Case (2) : When a ball is thrown at some angle ‘’ in the direction of motion
of the truck horizontal & vertical component of ball’s velocity w
...
t
...
Horizontal &
vertical component of ball’s velocity w
...
t
...


Case (3) : When a ball is thrown at some angle ‘’ in the opposite direction
of motion of the truck horizontal & vertical component of ball’s velocity
w
...
t
...

Horizontal & vertical component of ball’s velocity w
...
t
...


38
...
r
...

observer A standing on the moving platform is ucos and usin respectively
...
r
...
observer B sitting on
the ground is ux = ucos and uy = usin + v respectively
...
r
...

observer A standing on the moving platform is ucos and usin respectively
...
r
...
observer B sitting on
the ground is ux = ucos and uy = usin – v respectively
...


Projectile Motion on an Inclined Plane:Let a particle be projected up with a speed u from an inclined plane which
makes an angle  with the horizontal velocity of projection makes an angle
 with the inclined plane
...


Hence the component of initial velocity parallel and perpendicular to the
plane are equal to ucos and usinrespectively
...
e
...
e
...

Time of flight :
We know for oblique projectile motion T 
or we can say T 

2u sin 
g

2u
a

 Time of flight on an inclined plane T 

2u sin 
g cos 

Projectile Motion

Maximum height :
We know for oblique projectile motion H 
or we can say H 

u2 sin2 
2g

u2
2a 

40
...


Projectile Motion

(iii)

u2
g(1  sin )

EXAMPLE

Q1
Sol:

A particle will project with 20 m/s at angle 53° with horizontal find out time
at which its velocity will make an angle 45° will horizontal
...


Sol:

R= u

Projectile Motion

40 = u

2h
g
2  80
10

40 = 4 u
u = 10

42
...
Find out height
of the tower
...


Sol:


| v |  40
 = 30°

43
...
Find out change in momentum from t = 0 to t
...
It speed of bullet at the instant of firing is 500 m/sec
...


Q7

Particle will project for parabolic path then find out relation between elevation angle at top most point observed from projection point and angle of
projection
...
[i
...
magnitude of velocity]


i
...
component of a along v  
 
non consider
a v 
av
ˆ
v

,




only magnitude  v |v|
 v 



23
|a| 

10

5
10



5
m / s2
2

45
...
Find out ratio of range for both if ratio
of minimum kinetic energy is 1 : 2 and ratio of Hmax is 3 : 1
...
cos 
Rµ

Kmin  Hmax

R

R

1 3

2 1

R1
R

Projectile Motion

(i)



Kmin µ cos2 
cos µ

(ii)

Kmin

Hmax µ sin2 
sin   Hmax

3
2

46
...


Projectile Motion

48
...




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