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 Page 1


14. Vertical velocity of balloon (+ bag)
= ´ 12
5
18
 m/s
            =
10
3
 m/s
Horizontal velocity of balloon (+ bag)
= Wind velocity = = ´ 20 20
5
18
km h / m/s
=
50
9
 m/s
  = 5.55 m/s
\ tan a =
12
20
. i.e., sin a = 0.51
Bag is released at point A.
Let t be time, the bag takes from A to
reach ground.
Using, s ut at = +
1
2
2
( ) sin ( ) - =
æ
è
ç
ö
ø
÷ + - 50
10
3
1
2
2
a t g t
i.e., 5 50 0
2
t t - - = 1.7
\ t =
+ - - ´ ´ -
´
1.7 1.7 4 ( ) ( )
2
5 50
2 5
= 3.37 s
Vertical velocity of bag when it strikes
ground
   v
B
= - +
10
3
10 ( )( ) 3.37
=37.03 m/s
           v
w
=5.55 m/s 
\  Velocity of bag with which it strikes
ground
v v v
B net
= +
2 2
w
= 37.44 m/s
15. T
u
g
=
- 2 sin ( )
cos
a b
b
  =
´ ´ ° - °
°
2 20 2 45 30
10 30
sin( )
cos
       =1.69 s
     R
u
g
=
- 2
2
2
sin( )cos
cos
a b a
b
       =
´ ´ ° ´ °
°
2 20 2 15 45
10 30
2
2
( ) sin cos
cos
       =39 m
16.  T
u
g
=
+ 2 sin( )
cos
a b
b
    =
´ ´ ° + °
°
2 20 2 45 30
10 30
sin( )
cos
    =6.31 s
    R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
=
°
° + ° + °
( )
cos
[sin ( ) sin ]
20 2
10 30
90 30 30
2
2
= 145.71 m
17.   T
u
g
=
+ 2 sin( )
cos
a b
b
     =
2u
g
sin
cos
b
b
(Q a = ° 0 )
     =
2u
g
tanb
     =
´
°
2 20
10
30 tan
     =2.31 s
R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
     =
u
g
2
2
2 ( sin )
cos
b
b
[as a = ° 0 ]
     =
uT
cosb
 =
´
°
20
30
2.31
cos
          = 53.33 m
18. R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
  = + - +
u
g
2
2
2
cos
[sin{ ( ) } sin ]
b
a b b b
  = - +
u
g
2
2
cos
[sin( ) sin ]
q
p q q
[Q ( ) a b
p
+ =
2
]
or R
u
g
=
2
2
tanq q sec
60 | Mechanics-1
a
a
O
50 m
A
20 km/h v =
w
12 km/h
Page 2


14. Vertical velocity of balloon (+ bag)
= ´ 12
5
18
 m/s
            =
10
3
 m/s
Horizontal velocity of balloon (+ bag)
= Wind velocity = = ´ 20 20
5
18
km h / m/s
=
50
9
 m/s
  = 5.55 m/s
\ tan a =
12
20
. i.e., sin a = 0.51
Bag is released at point A.
Let t be time, the bag takes from A to
reach ground.
Using, s ut at = +
1
2
2
( ) sin ( ) - =
æ
è
ç
ö
ø
÷ + - 50
10
3
1
2
2
a t g t
i.e., 5 50 0
2
t t - - = 1.7
\ t =
+ - - ´ ´ -
´
1.7 1.7 4 ( ) ( )
2
5 50
2 5
= 3.37 s
Vertical velocity of bag when it strikes
ground
   v
B
= - +
10
3
10 ( )( ) 3.37
=37.03 m/s
           v
w
=5.55 m/s 
\  Velocity of bag with which it strikes
ground
v v v
B net
= +
2 2
w
= 37.44 m/s
15. T
u
g
=
- 2 sin ( )
cos
a b
b
  =
´ ´ ° - °
°
2 20 2 45 30
10 30
sin( )
cos
       =1.69 s
     R
u
g
=
- 2
2
2
sin( )cos
cos
a b a
b
       =
´ ´ ° ´ °
°
2 20 2 15 45
10 30
2
2
( ) sin cos
cos
       =39 m
16.  T
u
g
=
+ 2 sin( )
cos
a b
b
    =
´ ´ ° + °
°
2 20 2 45 30
10 30
sin( )
cos
    =6.31 s
    R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
=
°
° + ° + °
( )
cos
[sin ( ) sin ]
20 2
10 30
90 30 30
2
2
= 145.71 m
17.   T
u
g
=
+ 2 sin( )
cos
a b
b
     =
2u
g
sin
cos
b
b
(Q a = ° 0 )
     =
2u
g
tanb
     =
´
°
2 20
10
30 tan
     =2.31 s
R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
     =
u
g
2
2
2 ( sin )
cos
b
b
[as a = ° 0 ]
     =
uT
cosb
 =
´
°
20
30
2.31
cos
          = 53.33 m
18. R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
  = + - +
u
g
2
2
2
cos
[sin{ ( ) } sin ]
b
a b b b
  = - +
u
g
2
2
cos
[sin( ) sin ]
q
p q q
[Q ( ) a b
p
+ =
2
]
or R
u
g
=
2
2
tanq q sec
60 | Mechanics-1
a
a
O
50 m
A
20 km/h v =
w
12 km/h
19. (a)  Acceleration of particle 1 w.r.t. that of
particle 2
          = - - - ( ) ( ) g g
          =0
(b) Initial velocity of 1st particle = 20 j
^
 m/s
Initial velocity of 2nd particle 
= ° + ° ( cos sin )
^ ^
20 2 45 20 2 45 i j m/s
   = + ( )
^ ^
20 20 i j m/s
\ Initial velocity of 1st particle w.r.t. that
of 2nd particle
       = - + [( ) ( )]
^ ^ ^
20 20 20 j i j m/s
       = -20i
^
 m/s
       =20 m/s (downward)
(c) Horizontal velocity of 1st particle
      =0 m/s
Horizontal velocity of 2nd particle 
= 20 i
^
 m/s
\ Horizontal velocity of 1st particle w.r.t.
that of 2nd particle
= - 0 20 ( )
^
i
= - 20 i
^
 m/s
Relative displacement of 1st particle w.r.t. 
2nd particle at t = 2 s
= - ´ 20 2 i
^
  
= - 40 i
^
 m/s
\  Distance between the particles at t = 2 s
= 40 m
20. (a) As observed by passenger 
Vertical acceleration of stone
= - = g g 0
Horizontal velocity of stone
= - = v v 0
\ Path of the stone will be a straight line
(downwards).
(b)As observed by man standing on
ground 
Vertical acceleration of stone = g
Horizontal velocity of stone = v
\ Path of the stone will be parabolic.
21. (a) g g a
eff
= - - ( )
= + g a
= + 10 1
= 11 m/s
2
T
u
g
=
2 sin q
eff
=
´ ´ ° 2 2 30
11
sin
= 0.18 s
(b)Dotted path [(in lift) acceleration
upwards]
Full line path [In lift at rest or moving
with constant velocity upwards or
downwards].
(c)If lift is moving downward with
acceleration g.
g g g
eff
= - =0
22. Horizontal motion :
x u
1 1 1
= cos q and x u
2 2
= cos q
\ u u
1 1 2 2
20 cos cos q q + = …(i)
Projectile Motion 61
q
u
2
a m/s w m/s
q
u
q
Path of particle
u
Page 3


14. Vertical velocity of balloon (+ bag)
= ´ 12
5
18
 m/s
            =
10
3
 m/s
Horizontal velocity of balloon (+ bag)
= Wind velocity = = ´ 20 20
5
18
km h / m/s
=
50
9
 m/s
  = 5.55 m/s
\ tan a =
12
20
. i.e., sin a = 0.51
Bag is released at point A.
Let t be time, the bag takes from A to
reach ground.
Using, s ut at = +
1
2
2
( ) sin ( ) - =
æ
è
ç
ö
ø
÷ + - 50
10
3
1
2
2
a t g t
i.e., 5 50 0
2
t t - - = 1.7
\ t =
+ - - ´ ´ -
´
1.7 1.7 4 ( ) ( )
2
5 50
2 5
= 3.37 s
Vertical velocity of bag when it strikes
ground
   v
B
= - +
10
3
10 ( )( ) 3.37
=37.03 m/s
           v
w
=5.55 m/s 
\  Velocity of bag with which it strikes
ground
v v v
B net
= +
2 2
w
= 37.44 m/s
15. T
u
g
=
- 2 sin ( )
cos
a b
b
  =
´ ´ ° - °
°
2 20 2 45 30
10 30
sin( )
cos
       =1.69 s
     R
u
g
=
- 2
2
2
sin( )cos
cos
a b a
b
       =
´ ´ ° ´ °
°
2 20 2 15 45
10 30
2
2
( ) sin cos
cos
       =39 m
16.  T
u
g
=
+ 2 sin( )
cos
a b
b
    =
´ ´ ° + °
°
2 20 2 45 30
10 30
sin( )
cos
    =6.31 s
    R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
=
°
° + ° + °
( )
cos
[sin ( ) sin ]
20 2
10 30
90 30 30
2
2
= 145.71 m
17.   T
u
g
=
+ 2 sin( )
cos
a b
b
     =
2u
g
sin
cos
b
b
(Q a = ° 0 )
     =
2u
g
tanb
     =
´
°
2 20
10
30 tan
     =2.31 s
R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
     =
u
g
2
2
2 ( sin )
cos
b
b
[as a = ° 0 ]
     =
uT
cosb
 =
´
°
20
30
2.31
cos
          = 53.33 m
18. R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
  = + - +
u
g
2
2
2
cos
[sin{ ( ) } sin ]
b
a b b b
  = - +
u
g
2
2
cos
[sin( ) sin ]
q
p q q
[Q ( ) a b
p
+ =
2
]
or R
u
g
=
2
2
tanq q sec
60 | Mechanics-1
a
a
O
50 m
A
20 km/h v =
w
12 km/h
19. (a)  Acceleration of particle 1 w.r.t. that of
particle 2
          = - - - ( ) ( ) g g
          =0
(b) Initial velocity of 1st particle = 20 j
^
 m/s
Initial velocity of 2nd particle 
= ° + ° ( cos sin )
^ ^
20 2 45 20 2 45 i j m/s
   = + ( )
^ ^
20 20 i j m/s
\ Initial velocity of 1st particle w.r.t. that
of 2nd particle
       = - + [( ) ( )]
^ ^ ^
20 20 20 j i j m/s
       = -20i
^
 m/s
       =20 m/s (downward)
(c) Horizontal velocity of 1st particle
      =0 m/s
Horizontal velocity of 2nd particle 
= 20 i
^
 m/s
\ Horizontal velocity of 1st particle w.r.t.
that of 2nd particle
= - 0 20 ( )
^
i
= - 20 i
^
 m/s
Relative displacement of 1st particle w.r.t. 
2nd particle at t = 2 s
= - ´ 20 2 i
^
  
= - 40 i
^
 m/s
\  Distance between the particles at t = 2 s
= 40 m
20. (a) As observed by passenger 
Vertical acceleration of stone
= - = g g 0
Horizontal velocity of stone
= - = v v 0
\ Path of the stone will be a straight line
(downwards).
(b)As observed by man standing on
ground 
Vertical acceleration of stone = g
Horizontal velocity of stone = v
\ Path of the stone will be parabolic.
21. (a) g g a
eff
= - - ( )
= + g a
= + 10 1
= 11 m/s
2
T
u
g
=
2 sin q
eff
=
´ ´ ° 2 2 30
11
sin
= 0.18 s
(b)Dotted path [(in lift) acceleration
upwards]
Full line path [In lift at rest or moving
with constant velocity upwards or
downwards].
(c)If lift is moving downward with
acceleration g.
g g g
eff
= - =0
22. Horizontal motion :
x u
1 1 1
= cos q and x u
2 2
= cos q
\ u u
1 1 2 2
20 cos cos q q + = …(i)
Projectile Motion 61
q
u
2
a m/s w m/s
q
u
q
Path of particle
u
Vertical motion :
20
1
2
30
1 1
2
+ + - = ( sin ) ( ) u t g t q
+ + - ( sin ) ( ) u t g t
2 2
2
1
2
q
or ( sin sin ) u u t
1 1 2 2
10 q q - = …(ii)
Objective Questions (Level 1)
1. v i j
®
= + 3 4
^ ^
 and F i j
®
= - 4 3
^ ^
v F i j i j
® ®
× = + × - ( ) ( )
^ ^ ^ ^
3 4 4 3
= - = 12 12 0
\ F v
® ®
^
Path of the particle is circular.
Option (c) is correct.
2. Projectile motion is uniformly accelerated
everywhere even at the highest point.
Option (a) correct.
Option (b) incorrect.
At the highest point acceleration is
perpendicular to velocity.
Option (c) incorrect.
3. For range to be maximum
q = ° 45
i.e.,
u
u
x
= ° = cos 45
1
2
or v
u
x
=
2
( Q v u
x x
= )
=
20
2
= 14.14 m/s
= 14 m/s (approx) 
Option (b) is correct.
4. H (maximum height) =
u
g
2 2
2
sin a
\  H
u
g
1
2 2
2
=
sin q
 and H
u
g
2
2 2
90
=
° - sin ( ) q
=
u
g
2 2
cos q
Thus, 
H
H
1
2
2
2
=
sin
cos
q
q
Option (c) is correct.
5. Equation to trajectory is
Y x
gx
u
= - tan
cos sin
q
q q
1
2
2
2
       Y x
x
R
= - tanq
2
Area =
ò
Y dx
R
0
     = -
æ
è
ç
ö
ø
÷
ò
x
x
R
dx
R
tanq
2
0
   A
x x
R
R
= -
é
ë
ê
ù
û
ú
2 3
0
2 3
tanq
     = -
R R
2 3
2 3
tanq
     = -
é
ë
ê
ù
û
ú
R
2
2
1
3
tanq
62 | Mechanics-1
Y
v
F
X
i
j
®
®
u
2
q
1
x
2
x
1
d = 20 m
20
1
u
1
q
2
t
Page 4


14. Vertical velocity of balloon (+ bag)
= ´ 12
5
18
 m/s
            =
10
3
 m/s
Horizontal velocity of balloon (+ bag)
= Wind velocity = = ´ 20 20
5
18
km h / m/s
=
50
9
 m/s
  = 5.55 m/s
\ tan a =
12
20
. i.e., sin a = 0.51
Bag is released at point A.
Let t be time, the bag takes from A to
reach ground.
Using, s ut at = +
1
2
2
( ) sin ( ) - =
æ
è
ç
ö
ø
÷ + - 50
10
3
1
2
2
a t g t
i.e., 5 50 0
2
t t - - = 1.7
\ t =
+ - - ´ ´ -
´
1.7 1.7 4 ( ) ( )
2
5 50
2 5
= 3.37 s
Vertical velocity of bag when it strikes
ground
   v
B
= - +
10
3
10 ( )( ) 3.37
=37.03 m/s
           v
w
=5.55 m/s 
\  Velocity of bag with which it strikes
ground
v v v
B net
= +
2 2
w
= 37.44 m/s
15. T
u
g
=
- 2 sin ( )
cos
a b
b
  =
´ ´ ° - °
°
2 20 2 45 30
10 30
sin( )
cos
       =1.69 s
     R
u
g
=
- 2
2
2
sin( )cos
cos
a b a
b
       =
´ ´ ° ´ °
°
2 20 2 15 45
10 30
2
2
( ) sin cos
cos
       =39 m
16.  T
u
g
=
+ 2 sin( )
cos
a b
b
    =
´ ´ ° + °
°
2 20 2 45 30
10 30
sin( )
cos
    =6.31 s
    R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
=
°
° + ° + °
( )
cos
[sin ( ) sin ]
20 2
10 30
90 30 30
2
2
= 145.71 m
17.   T
u
g
=
+ 2 sin( )
cos
a b
b
     =
2u
g
sin
cos
b
b
(Q a = ° 0 )
     =
2u
g
tanb
     =
´
°
2 20
10
30 tan
     =2.31 s
R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
     =
u
g
2
2
2 ( sin )
cos
b
b
[as a = ° 0 ]
     =
uT
cosb
 =
´
°
20
30
2.31
cos
          = 53.33 m
18. R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
  = + - +
u
g
2
2
2
cos
[sin{ ( ) } sin ]
b
a b b b
  = - +
u
g
2
2
cos
[sin( ) sin ]
q
p q q
[Q ( ) a b
p
+ =
2
]
or R
u
g
=
2
2
tanq q sec
60 | Mechanics-1
a
a
O
50 m
A
20 km/h v =
w
12 km/h
19. (a)  Acceleration of particle 1 w.r.t. that of
particle 2
          = - - - ( ) ( ) g g
          =0
(b) Initial velocity of 1st particle = 20 j
^
 m/s
Initial velocity of 2nd particle 
= ° + ° ( cos sin )
^ ^
20 2 45 20 2 45 i j m/s
   = + ( )
^ ^
20 20 i j m/s
\ Initial velocity of 1st particle w.r.t. that
of 2nd particle
       = - + [( ) ( )]
^ ^ ^
20 20 20 j i j m/s
       = -20i
^
 m/s
       =20 m/s (downward)
(c) Horizontal velocity of 1st particle
      =0 m/s
Horizontal velocity of 2nd particle 
= 20 i
^
 m/s
\ Horizontal velocity of 1st particle w.r.t.
that of 2nd particle
= - 0 20 ( )
^
i
= - 20 i
^
 m/s
Relative displacement of 1st particle w.r.t. 
2nd particle at t = 2 s
= - ´ 20 2 i
^
  
= - 40 i
^
 m/s
\  Distance between the particles at t = 2 s
= 40 m
20. (a) As observed by passenger 
Vertical acceleration of stone
= - = g g 0
Horizontal velocity of stone
= - = v v 0
\ Path of the stone will be a straight line
(downwards).
(b)As observed by man standing on
ground 
Vertical acceleration of stone = g
Horizontal velocity of stone = v
\ Path of the stone will be parabolic.
21. (a) g g a
eff
= - - ( )
= + g a
= + 10 1
= 11 m/s
2
T
u
g
=
2 sin q
eff
=
´ ´ ° 2 2 30
11
sin
= 0.18 s
(b)Dotted path [(in lift) acceleration
upwards]
Full line path [In lift at rest or moving
with constant velocity upwards or
downwards].
(c)If lift is moving downward with
acceleration g.
g g g
eff
= - =0
22. Horizontal motion :
x u
1 1 1
= cos q and x u
2 2
= cos q
\ u u
1 1 2 2
20 cos cos q q + = …(i)
Projectile Motion 61
q
u
2
a m/s w m/s
q
u
q
Path of particle
u
Vertical motion :
20
1
2
30
1 1
2
+ + - = ( sin ) ( ) u t g t q
+ + - ( sin ) ( ) u t g t
2 2
2
1
2
q
or ( sin sin ) u u t
1 1 2 2
10 q q - = …(ii)
Objective Questions (Level 1)
1. v i j
®
= + 3 4
^ ^
 and F i j
®
= - 4 3
^ ^
v F i j i j
® ®
× = + × - ( ) ( )
^ ^ ^ ^
3 4 4 3
= - = 12 12 0
\ F v
® ®
^
Path of the particle is circular.
Option (c) is correct.
2. Projectile motion is uniformly accelerated
everywhere even at the highest point.
Option (a) correct.
Option (b) incorrect.
At the highest point acceleration is
perpendicular to velocity.
Option (c) incorrect.
3. For range to be maximum
q = ° 45
i.e.,
u
u
x
= ° = cos 45
1
2
or v
u
x
=
2
( Q v u
x x
= )
=
20
2
= 14.14 m/s
= 14 m/s (approx) 
Option (b) is correct.
4. H (maximum height) =
u
g
2 2
2
sin a
\  H
u
g
1
2 2
2
=
sin q
 and H
u
g
2
2 2
90
=
° - sin ( ) q
=
u
g
2 2
cos q
Thus, 
H
H
1
2
2
2
=
sin
cos
q
q
Option (c) is correct.
5. Equation to trajectory is
Y x
gx
u
= - tan
cos sin
q
q q
1
2
2
2
       Y x
x
R
= - tanq
2
Area =
ò
Y dx
R
0
     = -
æ
è
ç
ö
ø
÷
ò
x
x
R
dx
R
tanq
2
0
   A
x x
R
R
= -
é
ë
ê
ù
û
ú
2 3
0
2 3
tanq
     = -
R R
2 3
2 3
tanq
     = -
é
ë
ê
ù
û
ú
R
2
2
1
3
tanq
62 | Mechanics-1
Y
v
F
X
i
j
®
®
u
2
q
1
x
2
x
1
d = 20 m
20
1
u
1
q
2
t
    = -
é
ë
ê
ù
û
ú
4
2
1
3
0
4 2 2
2
v
g
sin cos tan q q q
    = -
é
ë
ê
ù
û
ú
4
2 3
0
4
2
3 2 2
v
g
sin cos sin cos q q q q
 = -
2
3
3 2
0
4
2
2 2 2
v
g
[ sin cos sin cos ] q q q q
6. v u cos cos f= q
or v
u
=
°
°
cos
cos
60
30
or v
u
=
3
KE at B mv =
1
2
2
= ×
1
2 3
2
m
u
=
K
3
Q
1
2
2
mu K =
æ
è
ç
ö
ø
÷
\ Option (b) is correct.
7.
u
g
u
g
2 2
2 1
2
sin q
=
æ
è
ç
ö
ø
÷
or sin 2
1
2
q =
or 2 30 q = °
\ q = ° 15
Option (a) is correct.
8. T
u
g
1
2
=
sin q
\ T
u
g
2
2 90
=
° - sin ( ) q
            =
2u
g
cosq
Thus, TT
g
u
g
1 2
2
2 2
=
sin cos q q
or TT
R
g
1 2
2
=
Þ R gTT =
1
2
1 2
Option (d) is correct.
9. R
max
= 1.6 m
u
g
2
= 1.6
Þ u = 4 m/s
\ T
u
g
=
° 2 45 sin
=
4 2
10
Number of jumps = =
10 2 10 2
4 2 10 T /
= 25
\ Grass hopper would go 
= ´ 25 1.6 m i.e., 40 m.
Option (d) is correct.
10. |Av. velocity|=
|Displacement|
time
  =
æ
è
ç
ö
ø
÷
+
1
2 2
2
2
T
R
H
/
  =
æ
è
ç
ö
ø
÷
+
æ
è
ç
ö
ø
÷
1
2
2
2
2 2
2
u
g
u
g
u
g
sin
sin cos sin
q
q q q
  = + u cos
sin
2
2
4
q
q
  = +
u
2
1 3
2
cos q
Option (b) is correct.
11. d R u T
2 2 2 2
= +
       =
æ
è
ç
ö
ø
÷
+
° æ
è
ç
ö
ø
÷
u
g
u
u
g
2
2
2
2
2 45 sin
       = +
u
g
u
g
4
2
4
2
2
       =
3
4
2
u
g
Projectile Motion 63
q
u
A
f
v
b
90°
45°
u
2
R = u /g T
Ball
uT
a
T
Page 5


14. Vertical velocity of balloon (+ bag)
= ´ 12
5
18
 m/s
            =
10
3
 m/s
Horizontal velocity of balloon (+ bag)
= Wind velocity = = ´ 20 20
5
18
km h / m/s
=
50
9
 m/s
  = 5.55 m/s
\ tan a =
12
20
. i.e., sin a = 0.51
Bag is released at point A.
Let t be time, the bag takes from A to
reach ground.
Using, s ut at = +
1
2
2
( ) sin ( ) - =
æ
è
ç
ö
ø
÷ + - 50
10
3
1
2
2
a t g t
i.e., 5 50 0
2
t t - - = 1.7
\ t =
+ - - ´ ´ -
´
1.7 1.7 4 ( ) ( )
2
5 50
2 5
= 3.37 s
Vertical velocity of bag when it strikes
ground
   v
B
= - +
10
3
10 ( )( ) 3.37
=37.03 m/s
           v
w
=5.55 m/s 
\  Velocity of bag with which it strikes
ground
v v v
B net
= +
2 2
w
= 37.44 m/s
15. T
u
g
=
- 2 sin ( )
cos
a b
b
  =
´ ´ ° - °
°
2 20 2 45 30
10 30
sin( )
cos
       =1.69 s
     R
u
g
=
- 2
2
2
sin( )cos
cos
a b a
b
       =
´ ´ ° ´ °
°
2 20 2 15 45
10 30
2
2
( ) sin cos
cos
       =39 m
16.  T
u
g
=
+ 2 sin( )
cos
a b
b
    =
´ ´ ° + °
°
2 20 2 45 30
10 30
sin( )
cos
    =6.31 s
    R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
=
°
° + ° + °
( )
cos
[sin ( ) sin ]
20 2
10 30
90 30 30
2
2
= 145.71 m
17.   T
u
g
=
+ 2 sin( )
cos
a b
b
     =
2u
g
sin
cos
b
b
(Q a = ° 0 )
     =
2u
g
tanb
     =
´
°
2 20
10
30 tan
     =2.31 s
R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
     =
u
g
2
2
2 ( sin )
cos
b
b
[as a = ° 0 ]
     =
uT
cosb
 =
´
°
20
30
2.31
cos
          = 53.33 m
18. R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
  = + - +
u
g
2
2
2
cos
[sin{ ( ) } sin ]
b
a b b b
  = - +
u
g
2
2
cos
[sin( ) sin ]
q
p q q
[Q ( ) a b
p
+ =
2
]
or R
u
g
=
2
2
tanq q sec
60 | Mechanics-1
a
a
O
50 m
A
20 km/h v =
w
12 km/h
19. (a)  Acceleration of particle 1 w.r.t. that of
particle 2
          = - - - ( ) ( ) g g
          =0
(b) Initial velocity of 1st particle = 20 j
^
 m/s
Initial velocity of 2nd particle 
= ° + ° ( cos sin )
^ ^
20 2 45 20 2 45 i j m/s
   = + ( )
^ ^
20 20 i j m/s
\ Initial velocity of 1st particle w.r.t. that
of 2nd particle
       = - + [( ) ( )]
^ ^ ^
20 20 20 j i j m/s
       = -20i
^
 m/s
       =20 m/s (downward)
(c) Horizontal velocity of 1st particle
      =0 m/s
Horizontal velocity of 2nd particle 
= 20 i
^
 m/s
\ Horizontal velocity of 1st particle w.r.t.
that of 2nd particle
= - 0 20 ( )
^
i
= - 20 i
^
 m/s
Relative displacement of 1st particle w.r.t. 
2nd particle at t = 2 s
= - ´ 20 2 i
^
  
= - 40 i
^
 m/s
\  Distance between the particles at t = 2 s
= 40 m
20. (a) As observed by passenger 
Vertical acceleration of stone
= - = g g 0
Horizontal velocity of stone
= - = v v 0
\ Path of the stone will be a straight line
(downwards).
(b)As observed by man standing on
ground 
Vertical acceleration of stone = g
Horizontal velocity of stone = v
\ Path of the stone will be parabolic.
21. (a) g g a
eff
= - - ( )
= + g a
= + 10 1
= 11 m/s
2
T
u
g
=
2 sin q
eff
=
´ ´ ° 2 2 30
11
sin
= 0.18 s
(b)Dotted path [(in lift) acceleration
upwards]
Full line path [In lift at rest or moving
with constant velocity upwards or
downwards].
(c)If lift is moving downward with
acceleration g.
g g g
eff
= - =0
22. Horizontal motion :
x u
1 1 1
= cos q and x u
2 2
= cos q
\ u u
1 1 2 2
20 cos cos q q + = …(i)
Projectile Motion 61
q
u
2
a m/s w m/s
q
u
q
Path of particle
u
Vertical motion :
20
1
2
30
1 1
2
+ + - = ( sin ) ( ) u t g t q
+ + - ( sin ) ( ) u t g t
2 2
2
1
2
q
or ( sin sin ) u u t
1 1 2 2
10 q q - = …(ii)
Objective Questions (Level 1)
1. v i j
®
= + 3 4
^ ^
 and F i j
®
= - 4 3
^ ^
v F i j i j
® ®
× = + × - ( ) ( )
^ ^ ^ ^
3 4 4 3
= - = 12 12 0
\ F v
® ®
^
Path of the particle is circular.
Option (c) is correct.
2. Projectile motion is uniformly accelerated
everywhere even at the highest point.
Option (a) correct.
Option (b) incorrect.
At the highest point acceleration is
perpendicular to velocity.
Option (c) incorrect.
3. For range to be maximum
q = ° 45
i.e.,
u
u
x
= ° = cos 45
1
2
or v
u
x
=
2
( Q v u
x x
= )
=
20
2
= 14.14 m/s
= 14 m/s (approx) 
Option (b) is correct.
4. H (maximum height) =
u
g
2 2
2
sin a
\  H
u
g
1
2 2
2
=
sin q
 and H
u
g
2
2 2
90
=
° - sin ( ) q
=
u
g
2 2
cos q
Thus, 
H
H
1
2
2
2
=
sin
cos
q
q
Option (c) is correct.
5. Equation to trajectory is
Y x
gx
u
= - tan
cos sin
q
q q
1
2
2
2
       Y x
x
R
= - tanq
2
Area =
ò
Y dx
R
0
     = -
æ
è
ç
ö
ø
÷
ò
x
x
R
dx
R
tanq
2
0
   A
x x
R
R
= -
é
ë
ê
ù
û
ú
2 3
0
2 3
tanq
     = -
R R
2 3
2 3
tanq
     = -
é
ë
ê
ù
û
ú
R
2
2
1
3
tanq
62 | Mechanics-1
Y
v
F
X
i
j
®
®
u
2
q
1
x
2
x
1
d = 20 m
20
1
u
1
q
2
t
    = -
é
ë
ê
ù
û
ú
4
2
1
3
0
4 2 2
2
v
g
sin cos tan q q q
    = -
é
ë
ê
ù
û
ú
4
2 3
0
4
2
3 2 2
v
g
sin cos sin cos q q q q
 = -
2
3
3 2
0
4
2
2 2 2
v
g
[ sin cos sin cos ] q q q q
6. v u cos cos f= q
or v
u
=
°
°
cos
cos
60
30
or v
u
=
3
KE at B mv =
1
2
2
= ×
1
2 3
2
m
u
=
K
3
Q
1
2
2
mu K =
æ
è
ç
ö
ø
÷
\ Option (b) is correct.
7.
u
g
u
g
2 2
2 1
2
sin q
=
æ
è
ç
ö
ø
÷
or sin 2
1
2
q =
or 2 30 q = °
\ q = ° 15
Option (a) is correct.
8. T
u
g
1
2
=
sin q
\ T
u
g
2
2 90
=
° - sin ( ) q
            =
2u
g
cosq
Thus, TT
g
u
g
1 2
2
2 2
=
sin cos q q
or TT
R
g
1 2
2
=
Þ R gTT =
1
2
1 2
Option (d) is correct.
9. R
max
= 1.6 m
u
g
2
= 1.6
Þ u = 4 m/s
\ T
u
g
=
° 2 45 sin
=
4 2
10
Number of jumps = =
10 2 10 2
4 2 10 T /
= 25
\ Grass hopper would go 
= ´ 25 1.6 m i.e., 40 m.
Option (d) is correct.
10. |Av. velocity|=
|Displacement|
time
  =
æ
è
ç
ö
ø
÷
+
1
2 2
2
2
T
R
H
/
  =
æ
è
ç
ö
ø
÷
+
æ
è
ç
ö
ø
÷
1
2
2
2
2 2
2
u
g
u
g
u
g
sin
sin cos sin
q
q q q
  = + u cos
sin
2
2
4
q
q
  = +
u
2
1 3
2
cos q
Option (b) is correct.
11. d R u T
2 2 2 2
= +
       =
æ
è
ç
ö
ø
÷
+
° æ
è
ç
ö
ø
÷
u
g
u
u
g
2
2
2
2
2 45 sin
       = +
u
g
u
g
4
2
4
2
2
       =
3
4
2
u
g
Projectile Motion 63
q
u
A
f
v
b
90°
45°
u
2
R = u /g T
Ball
uT
a
T
\               d
u
g
=
2
3 =
30
10
3
2
 = 90 3 m
Option (b) is correct.
12. At maximum height
dy
dt
= 0
i.e.,     
d
dt
t t ( ) 10 0
2
- =
or          10 2 0 - = t
or   t =5 s
\ Maximum height attained = - 10 5 5
2
( )
= 25 m
Option (d) is correct.
13. R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
As a = 0 (according to question)
      R
u
g
=
2
2
2
cos
[ sin ]
b
b
        =
´ °
´ °
( ) sin
cos
50 2 30
10 30
2
2
        =
1000
3
 m
Option (b) is correct.
14. First particle :
       H
max
=102 m
\     
u
g
2 2
2
102
sin a
=
Þ        u
g
2
2
102 2
60
=
´
° sin
[As a
p
= = °
3
60 ]
Second particle :
Range of the second particle will be equal
to that of particle 
if,     
u
g
u
g
2 2
2 2 sin sin f
=
a
          sin sin 2 3 f = a
            2 2 f = - p a
or           f = -
p
a
2
              = -
p p
2 3
 = = °
p
6
30
\ Maximum height attained by second
particle
               =
f u
g
2 2
2
sin
               =
´
°
´
° 102 2
60
30
2
2
2
g
g sin
sin
               =
´ 102 1 4
3 4
/
/
 =34 m
Option (d) is correct.
15. s ut at = +
1
2
2
\ ( ) ( sin ) ( ) - = ° + - 70 50 30
1
2
10
2
t t
or 5 25 70 0
2
t t - - =
or    t t
2
5 14 0 - - =
  ( ) ( ) t t - + = 7 2 0
t = 7 s
(- 2 s not possible)
Option (c) is correct.
16. Initial separation
         x R T = +5
= + ( cos ) u T T a 5
= + T u [( cos ) ] a 5
       = +
2
5
u
g
u
sin
[( cos ) ]
a
a
       =
´ ´
´
æ
è
ç
ö
ø
÷ +
é
ë
ê
ù
û
ú
2 80
4
5
10
80
3
5
5 =256 m
Option (d) is correct.
64 | Mechanics-1
53°
T
5T R
x
b
a
u
a = –g
R
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FAQs on DC Pandey Solutions: Projectile Motion- 2 - 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 involves both horizontal and vertical motion.
2. How is the trajectory of a projectile determined?
Ans. The trajectory of a projectile is determined by its initial velocity and the angle at which it is launched. The path is a parabolic curve, and its shape and range depend on these factors.
3. How can we calculate the maximum height reached by a projectile?
Ans. To calculate the maximum height reached by a projectile, we can use the formula: Maximum height (H) = (v^2 * sin^2θ)/(2g), where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.
4. What is the range of a projectile?
Ans. The range of a projectile is the horizontal distance covered by it before hitting the ground. It can be calculated using the formula: Range (R) = (v^2 * sin2θ)/g, where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.
5. How does air resistance affect projectile motion?
Ans. Air resistance can affect projectile motion by slowing down the projectile and altering its trajectory. In the absence of air resistance, the motion follows a perfect parabolic path. However, in reality, air resistance causes a decrease in the range and height of the projectile.
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