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 Page 1


Introductory Exercise 4.1
1. A particle projected at any angle with
horizontal will always move in a plane
and thus projectile motion is a
2-dimensional motion. The statement is
thus false.
2. At high speed the projectile may go to a
place where acceleration due to gravity
has some different value and as such the
motion may not be uniform accelerated.
The statement is thus true.
3. See article 4.1.
4. u = 40 2 m/s, q = ° 45
As horizontal acceleration would be zero.
v u u
x x
= = = cos q 40 m/s
 s u t u t
x x
= = = ( cos ) q 80 m
A : position of particle at time = 0.
B : position of particle at time = t.
As vertical acceleration would be - g
          v u gt
y y
= -
= = u gt sinq
             = - 40 20
             =20 m/s
s u t gt
y y
= × -
1
2
2
  
= - ´ ´ = 80
1
2
10 2 60
2
 m
\ v v v
x y
= +
2 2
= + 40 20
2 2
= 20 5 m/s
tan f = =
v
v
y
x
20
40
i.e., f =
æ
è
ç
ö
ø
÷
-
tan
1
1
2
s s s
x y
= +
2 2
= + 80 60
2 2
= 100 m
tan a =
s
s
y
x
=
60
80
i.e., a =
æ
è
ç
ö
ø
÷
-
tan
1
3
4
5. s u t g t
y y
= + -
1
2
2
( )
or s u t gt
y
= - ( sin ) q
1
2
2
or 15 20
1
2
2
= - t gt
or t t
2
4 3 0 = + =
i.e., t = 1  s and 3 s
6. See figure to the answer to question no. 4.
u = 40 m/s, q = ° 60
\ u
x
= ° = 40 60 20 cos m/s
Projectile Motion
4
s
y
g
f
v
v
y
v
x
u
x
s
x
A
a
u
u
y
B
Page 2


Introductory Exercise 4.1
1. A particle projected at any angle with
horizontal will always move in a plane
and thus projectile motion is a
2-dimensional motion. The statement is
thus false.
2. At high speed the projectile may go to a
place where acceleration due to gravity
has some different value and as such the
motion may not be uniform accelerated.
The statement is thus true.
3. See article 4.1.
4. u = 40 2 m/s, q = ° 45
As horizontal acceleration would be zero.
v u u
x x
= = = cos q 40 m/s
 s u t u t
x x
= = = ( cos ) q 80 m
A : position of particle at time = 0.
B : position of particle at time = t.
As vertical acceleration would be - g
          v u gt
y y
= -
= = u gt sinq
             = - 40 20
             =20 m/s
s u t gt
y y
= × -
1
2
2
  
= - ´ ´ = 80
1
2
10 2 60
2
 m
\ v v v
x y
= +
2 2
= + 40 20
2 2
= 20 5 m/s
tan f = =
v
v
y
x
20
40
i.e., f =
æ
è
ç
ö
ø
÷
-
tan
1
1
2
s s s
x y
= +
2 2
= + 80 60
2 2
= 100 m
tan a =
s
s
y
x
=
60
80
i.e., a =
æ
è
ç
ö
ø
÷
-
tan
1
3
4
5. s u t g t
y y
= + -
1
2
2
( )
or s u t gt
y
= - ( sin ) q
1
2
2
or 15 20
1
2
2
= - t gt
or t t
2
4 3 0 = + =
i.e., t = 1  s and 3 s
6. See figure to the answer to question no. 4.
u = 40 m/s, q = ° 60
\ u
x
= ° = 40 60 20 cos m/s
Projectile Motion
4
s
y
g
f
v
v
y
v
x
u
x
s
x
A
a
u
u
y
B
Thus, v u
x x
= = 20 m/s
As f = ° 45
tan f = =
v
v
y
x
1
\ y v
y x
= = 20 m/s
Thus, v v v v
x y x
= + =
2 2
2
= 20 2 m/s
Before reaching highest point
v u g t
y y
= + - ( )
\ 20 40 60 10 = ° - sin t
or 2 2 3 = - t
Þ t = - 2 3 1 ( ) s
After attaining highest point
- = - 20 40 3 10t
i.e., - = - 2 2 3 t
or t = + 2 3 1 ( ) s
7. Average velocity =
Range
Time of flight
=
u g
u g
2
2
sin 2a
a /
/
sin
= ´
2
2
2
u
g
g
u
sin cos
sin
a a
a
= ucos a
8. Change in velocity
          = - - + ( sin ) ( sin ) u u a a
          = -2usina
          =2usina (downward)
9. Formulae for R T , and H
max
 will be same
if the projection point and the point where
the particle lands are same and lie on a
horizontal line.
10. r i j
®
= + - [ ( ) ]
^ ^
3 4 5
2
t t t m
y-coordinate will be zero when
4 5 0
2
t t - =
i.e., t = 0 s, 
4
5
 s
t = 0 belongs to the initial point of
projection of the particle.
\ r i
®
= 0
^
 m
i.e., x = 0 m
At t = 0.8 s,
r i
®
= 2.4
^
 m
i.e., x = 2.4 m
11. v u
x x
= = 10 m/s
      =10 m/s
        v u
x x
= = 10 m/s
        v v
x
= f cos
Þ    cosf=
v
v
x
          =
10
v
          =
10
10
[as, v =
20
2
 (given)]
          =1
i.e.,       f = ° 0
\ Speed will be half of its initial value at
the highest point where f = ° 0 .
Thus,       t
u
g
=
sin q
 =
°
=
20 60
10
3
sin
 s
52 | Mechanics-1
a
a
u – u sin a
u sin a
t = T
u cos a
+ u sin a u
60°
u = 20 m/s
v
x
v
f
u = 20 cos 60°
x
Page 3


Introductory Exercise 4.1
1. A particle projected at any angle with
horizontal will always move in a plane
and thus projectile motion is a
2-dimensional motion. The statement is
thus false.
2. At high speed the projectile may go to a
place where acceleration due to gravity
has some different value and as such the
motion may not be uniform accelerated.
The statement is thus true.
3. See article 4.1.
4. u = 40 2 m/s, q = ° 45
As horizontal acceleration would be zero.
v u u
x x
= = = cos q 40 m/s
 s u t u t
x x
= = = ( cos ) q 80 m
A : position of particle at time = 0.
B : position of particle at time = t.
As vertical acceleration would be - g
          v u gt
y y
= -
= = u gt sinq
             = - 40 20
             =20 m/s
s u t gt
y y
= × -
1
2
2
  
= - ´ ´ = 80
1
2
10 2 60
2
 m
\ v v v
x y
= +
2 2
= + 40 20
2 2
= 20 5 m/s
tan f = =
v
v
y
x
20
40
i.e., f =
æ
è
ç
ö
ø
÷
-
tan
1
1
2
s s s
x y
= +
2 2
= + 80 60
2 2
= 100 m
tan a =
s
s
y
x
=
60
80
i.e., a =
æ
è
ç
ö
ø
÷
-
tan
1
3
4
5. s u t g t
y y
= + -
1
2
2
( )
or s u t gt
y
= - ( sin ) q
1
2
2
or 15 20
1
2
2
= - t gt
or t t
2
4 3 0 = + =
i.e., t = 1  s and 3 s
6. See figure to the answer to question no. 4.
u = 40 m/s, q = ° 60
\ u
x
= ° = 40 60 20 cos m/s
Projectile Motion
4
s
y
g
f
v
v
y
v
x
u
x
s
x
A
a
u
u
y
B
Thus, v u
x x
= = 20 m/s
As f = ° 45
tan f = =
v
v
y
x
1
\ y v
y x
= = 20 m/s
Thus, v v v v
x y x
= + =
2 2
2
= 20 2 m/s
Before reaching highest point
v u g t
y y
= + - ( )
\ 20 40 60 10 = ° - sin t
or 2 2 3 = - t
Þ t = - 2 3 1 ( ) s
After attaining highest point
- = - 20 40 3 10t
i.e., - = - 2 2 3 t
or t = + 2 3 1 ( ) s
7. Average velocity =
Range
Time of flight
=
u g
u g
2
2
sin 2a
a /
/
sin
= ´
2
2
2
u
g
g
u
sin cos
sin
a a
a
= ucos a
8. Change in velocity
          = - - + ( sin ) ( sin ) u u a a
          = -2usina
          =2usina (downward)
9. Formulae for R T , and H
max
 will be same
if the projection point and the point where
the particle lands are same and lie on a
horizontal line.
10. r i j
®
= + - [ ( ) ]
^ ^
3 4 5
2
t t t m
y-coordinate will be zero when
4 5 0
2
t t - =
i.e., t = 0 s, 
4
5
 s
t = 0 belongs to the initial point of
projection of the particle.
\ r i
®
= 0
^
 m
i.e., x = 0 m
At t = 0.8 s,
r i
®
= 2.4
^
 m
i.e., x = 2.4 m
11. v u
x x
= = 10 m/s
      =10 m/s
        v u
x x
= = 10 m/s
        v v
x
= f cos
Þ    cosf=
v
v
x
          =
10
v
          =
10
10
[as, v =
20
2
 (given)]
          =1
i.e.,       f = ° 0
\ Speed will be half of its initial value at
the highest point where f = ° 0 .
Thus,       t
u
g
=
sin q
 =
°
=
20 60
10
3
sin
 s
52 | Mechanics-1
a
a
u – u sin a
u sin a
t = T
u cos a
+ u sin a u
60°
u = 20 m/s
v
x
v
f
u = 20 cos 60°
x
Introductory Exercise 4.2
1. Time of flight
T
u
g
=
- 2 sin ( )
cos
a b
b
                =
´ ° - °
°
( sin( )
cos
2 10 60 30
10 30
                          =
2
3
 s
Using,             v u at = +
  v u u
x x
= = ° cos 60
    = ´ = 10
1
2
5 m/s
v u g T
y y
= + - ( )
       = ° - u gT sin 60
        = - 10
3
2
10
2
3
 =
-
æ
è
ç
ö
ø
÷
10
3 4
2 3
= -
5
3
 m/s
                   v v v
x y net
= +
2 2
      = +
-
æ
è
ç
ö
ø
÷
5
5
3
2
2
= + 5 1
1
3
  
=
10
3
 m/s  
2. Component of velocity perpendicular to
plane
           =v
net
cosb
           = ´ °
10
3
30 cos
           = ´
10
3
3
2
           =5 m/s
3. Let the particle collide at time t.
x u t
1
= ( cos ) q
and x v t
2
=
\ d x x = -
2 1
= + ( cos ) v u t q
= + ° [ cos ] 10 10 2 45 t = 20 t
Using equation, s ut at = +
1
2
2
For vertical motion of particle 1 :
h u t g t - = + - 10
1
2
2
( sin ) ( ) q
i.e., h u t gt = + - 10
1
2
2
( sin ) q …(i)
or h t gt - + - 10 10
1
2
2
For the vertical motion of particle 2 :
20
1
2
2
- = h gt
i.e., h gt = - 20
1
2
2
…(ii)
Comparing Eqs. (i) and (ii),
10 10
1
2
20
1
2
2 2
+ - = - t g t g t
Þ t = 1 s
\ d = 20 m
4.                          u = 10 m/s
v = 5 2 m/s
Projectile Motion 53
h
v
u
q
x
2
x
1
d
10 m
2
1
q f
B A
d
v
u
a
b
u
a = 60°
b = 30°
2
u = 10 m/s
2
g = 10 m/s
v
x
v
net
v
y
Page 4


Introductory Exercise 4.1
1. A particle projected at any angle with
horizontal will always move in a plane
and thus projectile motion is a
2-dimensional motion. The statement is
thus false.
2. At high speed the projectile may go to a
place where acceleration due to gravity
has some different value and as such the
motion may not be uniform accelerated.
The statement is thus true.
3. See article 4.1.
4. u = 40 2 m/s, q = ° 45
As horizontal acceleration would be zero.
v u u
x x
= = = cos q 40 m/s
 s u t u t
x x
= = = ( cos ) q 80 m
A : position of particle at time = 0.
B : position of particle at time = t.
As vertical acceleration would be - g
          v u gt
y y
= -
= = u gt sinq
             = - 40 20
             =20 m/s
s u t gt
y y
= × -
1
2
2
  
= - ´ ´ = 80
1
2
10 2 60
2
 m
\ v v v
x y
= +
2 2
= + 40 20
2 2
= 20 5 m/s
tan f = =
v
v
y
x
20
40
i.e., f =
æ
è
ç
ö
ø
÷
-
tan
1
1
2
s s s
x y
= +
2 2
= + 80 60
2 2
= 100 m
tan a =
s
s
y
x
=
60
80
i.e., a =
æ
è
ç
ö
ø
÷
-
tan
1
3
4
5. s u t g t
y y
= + -
1
2
2
( )
or s u t gt
y
= - ( sin ) q
1
2
2
or 15 20
1
2
2
= - t gt
or t t
2
4 3 0 = + =
i.e., t = 1  s and 3 s
6. See figure to the answer to question no. 4.
u = 40 m/s, q = ° 60
\ u
x
= ° = 40 60 20 cos m/s
Projectile Motion
4
s
y
g
f
v
v
y
v
x
u
x
s
x
A
a
u
u
y
B
Thus, v u
x x
= = 20 m/s
As f = ° 45
tan f = =
v
v
y
x
1
\ y v
y x
= = 20 m/s
Thus, v v v v
x y x
= + =
2 2
2
= 20 2 m/s
Before reaching highest point
v u g t
y y
= + - ( )
\ 20 40 60 10 = ° - sin t
or 2 2 3 = - t
Þ t = - 2 3 1 ( ) s
After attaining highest point
- = - 20 40 3 10t
i.e., - = - 2 2 3 t
or t = + 2 3 1 ( ) s
7. Average velocity =
Range
Time of flight
=
u g
u g
2
2
sin 2a
a /
/
sin
= ´
2
2
2
u
g
g
u
sin cos
sin
a a
a
= ucos a
8. Change in velocity
          = - - + ( sin ) ( sin ) u u a a
          = -2usina
          =2usina (downward)
9. Formulae for R T , and H
max
 will be same
if the projection point and the point where
the particle lands are same and lie on a
horizontal line.
10. r i j
®
= + - [ ( ) ]
^ ^
3 4 5
2
t t t m
y-coordinate will be zero when
4 5 0
2
t t - =
i.e., t = 0 s, 
4
5
 s
t = 0 belongs to the initial point of
projection of the particle.
\ r i
®
= 0
^
 m
i.e., x = 0 m
At t = 0.8 s,
r i
®
= 2.4
^
 m
i.e., x = 2.4 m
11. v u
x x
= = 10 m/s
      =10 m/s
        v u
x x
= = 10 m/s
        v v
x
= f cos
Þ    cosf=
v
v
x
          =
10
v
          =
10
10
[as, v =
20
2
 (given)]
          =1
i.e.,       f = ° 0
\ Speed will be half of its initial value at
the highest point where f = ° 0 .
Thus,       t
u
g
=
sin q
 =
°
=
20 60
10
3
sin
 s
52 | Mechanics-1
a
a
u – u sin a
u sin a
t = T
u cos a
+ u sin a u
60°
u = 20 m/s
v
x
v
f
u = 20 cos 60°
x
Introductory Exercise 4.2
1. Time of flight
T
u
g
=
- 2 sin ( )
cos
a b
b
                =
´ ° - °
°
( sin( )
cos
2 10 60 30
10 30
                          =
2
3
 s
Using,             v u at = +
  v u u
x x
= = ° cos 60
    = ´ = 10
1
2
5 m/s
v u g T
y y
= + - ( )
       = ° - u gT sin 60
        = - 10
3
2
10
2
3
 =
-
æ
è
ç
ö
ø
÷
10
3 4
2 3
= -
5
3
 m/s
                   v v v
x y net
= +
2 2
      = +
-
æ
è
ç
ö
ø
÷
5
5
3
2
2
= + 5 1
1
3
  
=
10
3
 m/s  
2. Component of velocity perpendicular to
plane
           =v
net
cosb
           = ´ °
10
3
30 cos
           = ´
10
3
3
2
           =5 m/s
3. Let the particle collide at time t.
x u t
1
= ( cos ) q
and x v t
2
=
\ d x x = -
2 1
= + ( cos ) v u t q
= + ° [ cos ] 10 10 2 45 t = 20 t
Using equation, s ut at = +
1
2
2
For vertical motion of particle 1 :
h u t g t - = + - 10
1
2
2
( sin ) ( ) q
i.e., h u t gt = + - 10
1
2
2
( sin ) q …(i)
or h t gt - + - 10 10
1
2
2
For the vertical motion of particle 2 :
20
1
2
2
- = h gt
i.e., h gt = - 20
1
2
2
…(ii)
Comparing Eqs. (i) and (ii),
10 10
1
2
20
1
2
2 2
+ - = - t g t g t
Þ t = 1 s
\ d = 20 m
4.                          u = 10 m/s
v = 5 2 m/s
Projectile Motion 53
h
v
u
q
x
2
x
1
d
10 m
2
1
q f
B A
d
v
u
a
b
u
a = 60°
b = 30°
2
u = 10 m/s
2
g = 10 m/s
v
x
v
net
v
y
q = ° 30
f = ° 45
d = 15 m
Let the particles meet (or are in the same
vertical time t).
\ d u t v t = + f ( cos ) ( cos ) q
Þ 15 10 30 5 2 45 = ° + ° ( cos cos ) t
or 15 5 3 5 = + ( ) t
or t =
+
3
3 1
 s
= 1.009 s
Now, let us find time of flight of A and B
T
u
g
A
=
2 sin q
= 1 s
As T t
A
< , particle A will touch ground
before the expected time t of collision.
\      Ans : NO.
5. For range to be maximum
  a
p b
= +
4 2
          = +
p p
4
6
2
/
=
p
3
   = ° 60
6. At point A velocity ( ) v
®
 of the particle will
be parallel to the inclined plane.
   u = 40 m/s
a = ° 60
b = ° 30
        g = 10 m/s
2
\ f = b
v u u
x x
= = cos a
v v v
x
= f = cos cos b
or u v cos cos a b =
Þ v
u
= =
°
°
cos
cos
cos
cos
a
b
40 60
30
=
æ
è
ç
ö
ø
÷ 40
1
2
3
2
 
=
40
3
 m/s
7. (a) At time t, vertical displacement of A
= Vertical displacement of B
      ( sin ) v t gt v t gt
A B
q - = -
1
2
1
2
2 2
i.e.,                v v
A B
sin q =
sin q =
v
v
B
A
            = =
10
20
1
2
\     q = ° 30
(b) x v t
A
= ( cos ) q
        = ° ´ ( cos ) 20 30
1
2
         = 5 3 m
54 | Mechanics-1
a
b = 30° =
u
p
6
a
b
u
A
f A
v
time t
v = 10 m/s
B
2
a = – 10 m/s
v = 20 m/s
A
A
B
x
Page 5


Introductory Exercise 4.1
1. A particle projected at any angle with
horizontal will always move in a plane
and thus projectile motion is a
2-dimensional motion. The statement is
thus false.
2. At high speed the projectile may go to a
place where acceleration due to gravity
has some different value and as such the
motion may not be uniform accelerated.
The statement is thus true.
3. See article 4.1.
4. u = 40 2 m/s, q = ° 45
As horizontal acceleration would be zero.
v u u
x x
= = = cos q 40 m/s
 s u t u t
x x
= = = ( cos ) q 80 m
A : position of particle at time = 0.
B : position of particle at time = t.
As vertical acceleration would be - g
          v u gt
y y
= -
= = u gt sinq
             = - 40 20
             =20 m/s
s u t gt
y y
= × -
1
2
2
  
= - ´ ´ = 80
1
2
10 2 60
2
 m
\ v v v
x y
= +
2 2
= + 40 20
2 2
= 20 5 m/s
tan f = =
v
v
y
x
20
40
i.e., f =
æ
è
ç
ö
ø
÷
-
tan
1
1
2
s s s
x y
= +
2 2
= + 80 60
2 2
= 100 m
tan a =
s
s
y
x
=
60
80
i.e., a =
æ
è
ç
ö
ø
÷
-
tan
1
3
4
5. s u t g t
y y
= + -
1
2
2
( )
or s u t gt
y
= - ( sin ) q
1
2
2
or 15 20
1
2
2
= - t gt
or t t
2
4 3 0 = + =
i.e., t = 1  s and 3 s
6. See figure to the answer to question no. 4.
u = 40 m/s, q = ° 60
\ u
x
= ° = 40 60 20 cos m/s
Projectile Motion
4
s
y
g
f
v
v
y
v
x
u
x
s
x
A
a
u
u
y
B
Thus, v u
x x
= = 20 m/s
As f = ° 45
tan f = =
v
v
y
x
1
\ y v
y x
= = 20 m/s
Thus, v v v v
x y x
= + =
2 2
2
= 20 2 m/s
Before reaching highest point
v u g t
y y
= + - ( )
\ 20 40 60 10 = ° - sin t
or 2 2 3 = - t
Þ t = - 2 3 1 ( ) s
After attaining highest point
- = - 20 40 3 10t
i.e., - = - 2 2 3 t
or t = + 2 3 1 ( ) s
7. Average velocity =
Range
Time of flight
=
u g
u g
2
2
sin 2a
a /
/
sin
= ´
2
2
2
u
g
g
u
sin cos
sin
a a
a
= ucos a
8. Change in velocity
          = - - + ( sin ) ( sin ) u u a a
          = -2usina
          =2usina (downward)
9. Formulae for R T , and H
max
 will be same
if the projection point and the point where
the particle lands are same and lie on a
horizontal line.
10. r i j
®
= + - [ ( ) ]
^ ^
3 4 5
2
t t t m
y-coordinate will be zero when
4 5 0
2
t t - =
i.e., t = 0 s, 
4
5
 s
t = 0 belongs to the initial point of
projection of the particle.
\ r i
®
= 0
^
 m
i.e., x = 0 m
At t = 0.8 s,
r i
®
= 2.4
^
 m
i.e., x = 2.4 m
11. v u
x x
= = 10 m/s
      =10 m/s
        v u
x x
= = 10 m/s
        v v
x
= f cos
Þ    cosf=
v
v
x
          =
10
v
          =
10
10
[as, v =
20
2
 (given)]
          =1
i.e.,       f = ° 0
\ Speed will be half of its initial value at
the highest point where f = ° 0 .
Thus,       t
u
g
=
sin q
 =
°
=
20 60
10
3
sin
 s
52 | Mechanics-1
a
a
u – u sin a
u sin a
t = T
u cos a
+ u sin a u
60°
u = 20 m/s
v
x
v
f
u = 20 cos 60°
x
Introductory Exercise 4.2
1. Time of flight
T
u
g
=
- 2 sin ( )
cos
a b
b
                =
´ ° - °
°
( sin( )
cos
2 10 60 30
10 30
                          =
2
3
 s
Using,             v u at = +
  v u u
x x
= = ° cos 60
    = ´ = 10
1
2
5 m/s
v u g T
y y
= + - ( )
       = ° - u gT sin 60
        = - 10
3
2
10
2
3
 =
-
æ
è
ç
ö
ø
÷
10
3 4
2 3
= -
5
3
 m/s
                   v v v
x y net
= +
2 2
      = +
-
æ
è
ç
ö
ø
÷
5
5
3
2
2
= + 5 1
1
3
  
=
10
3
 m/s  
2. Component of velocity perpendicular to
plane
           =v
net
cosb
           = ´ °
10
3
30 cos
           = ´
10
3
3
2
           =5 m/s
3. Let the particle collide at time t.
x u t
1
= ( cos ) q
and x v t
2
=
\ d x x = -
2 1
= + ( cos ) v u t q
= + ° [ cos ] 10 10 2 45 t = 20 t
Using equation, s ut at = +
1
2
2
For vertical motion of particle 1 :
h u t g t - = + - 10
1
2
2
( sin ) ( ) q
i.e., h u t gt = + - 10
1
2
2
( sin ) q …(i)
or h t gt - + - 10 10
1
2
2
For the vertical motion of particle 2 :
20
1
2
2
- = h gt
i.e., h gt = - 20
1
2
2
…(ii)
Comparing Eqs. (i) and (ii),
10 10
1
2
20
1
2
2 2
+ - = - t g t g t
Þ t = 1 s
\ d = 20 m
4.                          u = 10 m/s
v = 5 2 m/s
Projectile Motion 53
h
v
u
q
x
2
x
1
d
10 m
2
1
q f
B A
d
v
u
a
b
u
a = 60°
b = 30°
2
u = 10 m/s
2
g = 10 m/s
v
x
v
net
v
y
q = ° 30
f = ° 45
d = 15 m
Let the particles meet (or are in the same
vertical time t).
\ d u t v t = + f ( cos ) ( cos ) q
Þ 15 10 30 5 2 45 = ° + ° ( cos cos ) t
or 15 5 3 5 = + ( ) t
or t =
+
3
3 1
 s
= 1.009 s
Now, let us find time of flight of A and B
T
u
g
A
=
2 sin q
= 1 s
As T t
A
< , particle A will touch ground
before the expected time t of collision.
\      Ans : NO.
5. For range to be maximum
  a
p b
= +
4 2
          = +
p p
4
6
2
/
=
p
3
   = ° 60
6. At point A velocity ( ) v
®
 of the particle will
be parallel to the inclined plane.
   u = 40 m/s
a = ° 60
b = ° 30
        g = 10 m/s
2
\ f = b
v u u
x x
= = cos a
v v v
x
= f = cos cos b
or u v cos cos a b =
Þ v
u
= =
°
°
cos
cos
cos
cos
a
b
40 60
30
=
æ
è
ç
ö
ø
÷ 40
1
2
3
2
 
=
40
3
 m/s
7. (a) At time t, vertical displacement of A
= Vertical displacement of B
      ( sin ) v t gt v t gt
A B
q - = -
1
2
1
2
2 2
i.e.,                v v
A B
sin q =
sin q =
v
v
B
A
            = =
10
20
1
2
\     q = ° 30
(b) x v t
A
= ( cos ) q
        = ° ´ ( cos ) 20 30
1
2
         = 5 3 m
54 | Mechanics-1
a
b = 30° =
u
p
6
a
b
u
A
f A
v
time t
v = 10 m/s
B
2
a = – 10 m/s
v = 20 m/s
A
A
B
x
AIEEE Cor ner
Subjective Questions (Level 1)
1. (a) R
u
g
= =
°
2 2
2 20 2 90
10
sin ( ) sin q
= 80 m
   H
u
g
= =
°
2 2 2 2
2
20 2 45
10
sin ( ) sin q
= 40 m
   T
u
g
= =
° 2 2 20 2 45
10
sin ( ) sin q
= 4 s
(b) u i j
®
= + ( )
^ ^
20 20 m/s and a j
®
= - 10
^
 m/s
2
- = +
® ® ®
v u a t
= + + - ( ) ( )
^ ^ ^
20 20 10 1 i j j (at t = 1 s)
= - ( )
^ ^
20 10 i j m/s
(c) Time of flight 
T
u
g
=
2 sin q
=
´ ´
æ
è
ç
ö
ø
÷
2 20 2
1
2
10
= 4 s
(c) \ Velocity of particle at the time of
collision with ground.
= + + - ( ) ( )
^ ^ ^
20 20 10 4 i j j
= - ( )
^ ^
20 20 i j m/s
2. (a) s ut at = +
1
2
2
\ ( ) ( ) ( ) - = + + - 40 20
1
2
10
2
T T
or 5 20 40 0
2
T T - - =
or   T T
2
4 8 0 - - =
\ T =
- - + - - - ( ) ( ) ( )( )
( )
4 4 4 1 8
2 1
2
Leaving - ive sign which is not positive.
                =
+ 4 48
2
 =
+ 4 4 3
2
  = + ( ) 2 2 3 s
              =5.46 s
       R = ´ = 20 5.46 109.2 m
(b) s ut at = +
1
2
2
( ) ( ) - = ´ + - 40
1
2
10
2
0 T T
Þ T = 2 2 s
= 2.83 s
R = ´ 2.83 20
= 56.6 m
(c) s ut at = +
1
2
2
( ) ( ) ( ) - = - + - 40 20
1
2
10
2
T T
or 5 20 40 0
2
T T + - =
or T T
2
4 8 0 + - =
\ T =
- + - - 4 4 4 1 8
2 1
2
( ) ( ) ( )
( )
              =
- + 4 4 3
2
 = - + 2 2 3 = 1.46 s
Projectile Motion 55
R
– 40 m
+
T
2
a = – 10 m/s
45°
20Ö2 m/s
20 m/s
R
– 40 m
+
T
20 m/s
2
a = – 10 m/s
R
T
– 40 m
20 Ö2 m/s
Read More
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FAQs on DC Pandey Solutions: Projectile Motion- 1 - DC Pandey Solutions for NEET Physics

1. What is projectile motion?
Ans. Projectile motion refers to the motion of an object that is launched into the air and moves along a curved path under the influence of gravity. It is characterized by two independent components: horizontal motion at a constant velocity and vertical motion under the acceleration due to gravity.
2. What are the key equations used in projectile motion?
Ans. The key equations used in projectile motion are: - Horizontal motion equation: x = v₀x * t - Vertical motion equation: y = v₀y * t - (1/2) * g * t² - Range equation: R = (v₀x * 2 * v₀y) / g - Time of flight equation: T = (2 * v₀y) / g - Maximum height equation: H = (v₀y)² / (2 * g) where x and y represent the horizontal and vertical displacements, v₀x and v₀y represent the initial horizontal and vertical velocities, g represents the acceleration due to gravity, t represents time, R represents the range, T represents the time of flight, and H represents the maximum height.
3. How does the angle of projection affect the range and maximum height in projectile motion?
Ans. The angle of projection affects both the range and maximum height in projectile motion. The range is maximum when the angle of projection is 45 degrees, as it provides an equal combination of horizontal and vertical velocities. On the other hand, the maximum height is achieved when the angle of projection is 90 degrees (vertical projection), as it maximizes the vertical velocity component.
4. How does the initial velocity affect the range and maximum height in projectile motion?
Ans. The initial velocity affects both the range and maximum height in projectile motion. The range is directly proportional to the square of the initial velocity. Thus, increasing the initial velocity will result in a greater range. However, the maximum height is directly proportional to the square of the initial vertical velocity component. Therefore, increasing the initial velocity will increase the maximum height as well.
5. What are the different types of projectile motion?
Ans. There are two main types of projectile motion: horizontal projectile motion and angled projectile motion. In horizontal projectile motion, the object is launched horizontally, resulting in a constant horizontal velocity and a vertical motion due to gravity. In angled projectile motion, the object is launched at an angle to the horizontal, resulting in both horizontal and vertical motions.
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