Chapter 4 - Projectile Motion (Part - 3) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

DC Pandey (Questions & Solutions) of Physics: NEET

Created by: Ciel Knowledge

NEET : Chapter 4 - Projectile Motion (Part - 3) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

 Page 1


2. t
d
u
= also h gt =
1
2
2
\ h g
d
u
= ×
1
2
2
2
Þ d
u h
g
2
2
2
=
Option (b) is correct.
3. R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
Substituting v = 10 m/s
           g =10 m/s
2
         a b + = ° 90
and        b = ° 30
R =
°
° + ° + °
( )
cos
[sin( ) sin ]
10
10 30
90 60 30
2
2
   =
æ
è
ç
ö
ø
÷
+
é
ë
ê
ù
û
ú
10
3
4
1
2
1
2
 =
40
3
 m 
Option (c) is correct.
4. y x g
x
u
= - tan
sin cos
q
q q
1
2
2
2
dy
dx
x
u
= - tan
sin cos
q
q q
1
2
2
2
Slope of trajectory = - tan
sin cos
q
q q
x
u
2
Substituting x u t = ( cos ) q
    
dy
dx
u t
u
= - tan
( cos )
sin cos
q
q
q q
2
  = - tan
sin
q
q
t
u
Option (a) is correct.
5.         y x =b
2
\        
dy
dt
x
dx
dt
= × b 2
and
d y
dt
x
d x
dt
dx
dt
2
2
2
2
2
2 = × +
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
b
or  a b = +
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
2
2
2
2
x
d x
dt
dx
dt
Now, as 
d x
dt
2
2
0 = 
(acceleration being along y-axis only)
dx
dt
=
a
b 2
Option (d) is correct.
6. As the projectile hits the inclined plane
horizontally
PQ H
u
g
= =
max
sin
2 2
2
a
and OQ
u
g
=
2
sin cos a a
\ Range on inclined plane
  R PQ OQ ¢ = + ( ) ( )
2 2
    =
æ
è
ç
ö
ø
÷
+
æ
è
ç
ö
ø
÷
u
g
u
g
2 2
2
2
2
2
sin sin cos a a a
    = +
u
g
2 4
2 2
4
sin
sin cos
a
a a
    =
æ
è
ç
ö
ø
÷
+
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷
u
g
2
4 2
2
1
4
3
4
3
2
1
2
Projectile Motion 67
Q
P
R'
O
u
a
b
x
Slope
t
Slope
T
tan q
O
T/2
T = Time of flight
h
u
d
time = t
Page 2


2. t
d
u
= also h gt =
1
2
2
\ h g
d
u
= ×
1
2
2
2
Þ d
u h
g
2
2
2
=
Option (b) is correct.
3. R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
Substituting v = 10 m/s
           g =10 m/s
2
         a b + = ° 90
and        b = ° 30
R =
°
° + ° + °
( )
cos
[sin( ) sin ]
10
10 30
90 60 30
2
2
   =
æ
è
ç
ö
ø
÷
+
é
ë
ê
ù
û
ú
10
3
4
1
2
1
2
 =
40
3
 m 
Option (c) is correct.
4. y x g
x
u
= - tan
sin cos
q
q q
1
2
2
2
dy
dx
x
u
= - tan
sin cos
q
q q
1
2
2
2
Slope of trajectory = - tan
sin cos
q
q q
x
u
2
Substituting x u t = ( cos ) q
    
dy
dx
u t
u
= - tan
( cos )
sin cos
q
q
q q
2
  = - tan
sin
q
q
t
u
Option (a) is correct.
5.         y x =b
2
\        
dy
dt
x
dx
dt
= × b 2
and
d y
dt
x
d x
dt
dx
dt
2
2
2
2
2
2 = × +
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
b
or  a b = +
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
2
2
2
2
x
d x
dt
dx
dt
Now, as 
d x
dt
2
2
0 = 
(acceleration being along y-axis only)
dx
dt
=
a
b 2
Option (d) is correct.
6. As the projectile hits the inclined plane
horizontally
PQ H
u
g
= =
max
sin
2 2
2
a
and OQ
u
g
=
2
sin cos a a
\ Range on inclined plane
  R PQ OQ ¢ = + ( ) ( )
2 2
    =
æ
è
ç
ö
ø
÷
+
æ
è
ç
ö
ø
÷
u
g
u
g
2 2
2
2
2
2
sin sin cos a a a
    = +
u
g
2 4
2 2
4
sin
sin cos
a
a a
    =
æ
è
ç
ö
ø
÷
+
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷
u
g
2
4 2
2
1
4
3
4
3
2
1
2
Projectile Motion 67
Q
P
R'
O
u
a
b
x
Slope
t
Slope
T
tan q
O
T/2
T = Time of flight
h
u
d
time = t
    = +
u
g
2
9
64
3
16
    =
u
g
2
21
8
Option (d) is correct.
7. Let the projectile hits the inclined plane at 
P at time t
PQ u gt = - ( sin ) a
1
2
2
Further, QR vt u t = - ( cos ) a
and tan b =
PQ
QR
i.e., OQ PQ tan b =
t v u gt t [ cos ] tan sin - = -
é
ë
ê
ù
û
ú
a b a
1
2
or      10 1 60 30 [ cos ] tan - ° °
      = ° - ´ ´
é
ë
ê
ù
û
ú
10 60
1
2
10 sin t
or          
5
3
5 3 5 = - t
or           t = - 3
1
3
                    =
2
3
 s
8. y y x
2
2 2 + + =
\ 2 2 y
dy
dt
dy
dt
dx
dt
+ =
or ( ) 2 2 y
dy
dt
dx
dt
+ =
or
dx
dt
y = + 10 1 ( )
or
d x
dt
dy
dt
2
2
10 =
æ
è
ç
ö
ø
÷
= 50 m/s
2
Acceleration of the particle
       =
d x
dt
2
2
       =50 m/s
2
[as 
dy
dt
= constant]
9. tan tan tan a q = + f
              = + tan tan q q = 2 tan q
\ q
a
=
æ
è
ç
ö
ø
÷
-
tan
tan
1
2
Option (c) is correct.
10. At any time t,
Horizontal distance between particles
x t = - + 20 3 10 3 10 3 ( )
= - 20 3 1 ( ) t
Vertical distance between particles at
time t
y t t = - = ( ) 30 10 20
Distance between particles at time t
        D x y = +
2 2
or D t t
2 2 2
20 3 1 20 = - + [ ( )] [ ]
      = - + 400 3 1 400
2 2
[ ( ) ] t t
For D to minimum
dD
dt
= 0
3 2 1 1 2 0 ´ - - + = ( )( ) t t
or 6 6 2 0 t t - + =
i.e., t =
3
4
 s
\ D
min
2
2 2
400 3 1
3
4
400
3
4
= -
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
+
æ
è
ç
ö
ø
÷
= + ´ 75 75 3
Þ D
min
= 10 3 m
Option (b) is correct.
68 | Mechanics-1
20Ö3m
20Ö3 m/s
20 m/s
30° 60°
q f
H
R/2 R/2
a
b
v
v = 10 Ö3 m/s
b
P (time = t)
R
Q
(u cos a) t
vt
u
R
a
u = 10 Ö3 m/s
i.e., v = u
Page 3


2. t
d
u
= also h gt =
1
2
2
\ h g
d
u
= ×
1
2
2
2
Þ d
u h
g
2
2
2
=
Option (b) is correct.
3. R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
Substituting v = 10 m/s
           g =10 m/s
2
         a b + = ° 90
and        b = ° 30
R =
°
° + ° + °
( )
cos
[sin( ) sin ]
10
10 30
90 60 30
2
2
   =
æ
è
ç
ö
ø
÷
+
é
ë
ê
ù
û
ú
10
3
4
1
2
1
2
 =
40
3
 m 
Option (c) is correct.
4. y x g
x
u
= - tan
sin cos
q
q q
1
2
2
2
dy
dx
x
u
= - tan
sin cos
q
q q
1
2
2
2
Slope of trajectory = - tan
sin cos
q
q q
x
u
2
Substituting x u t = ( cos ) q
    
dy
dx
u t
u
= - tan
( cos )
sin cos
q
q
q q
2
  = - tan
sin
q
q
t
u
Option (a) is correct.
5.         y x =b
2
\        
dy
dt
x
dx
dt
= × b 2
and
d y
dt
x
d x
dt
dx
dt
2
2
2
2
2
2 = × +
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
b
or  a b = +
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
2
2
2
2
x
d x
dt
dx
dt
Now, as 
d x
dt
2
2
0 = 
(acceleration being along y-axis only)
dx
dt
=
a
b 2
Option (d) is correct.
6. As the projectile hits the inclined plane
horizontally
PQ H
u
g
= =
max
sin
2 2
2
a
and OQ
u
g
=
2
sin cos a a
\ Range on inclined plane
  R PQ OQ ¢ = + ( ) ( )
2 2
    =
æ
è
ç
ö
ø
÷
+
æ
è
ç
ö
ø
÷
u
g
u
g
2 2
2
2
2
2
sin sin cos a a a
    = +
u
g
2 4
2 2
4
sin
sin cos
a
a a
    =
æ
è
ç
ö
ø
÷
+
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷
u
g
2
4 2
2
1
4
3
4
3
2
1
2
Projectile Motion 67
Q
P
R'
O
u
a
b
x
Slope
t
Slope
T
tan q
O
T/2
T = Time of flight
h
u
d
time = t
    = +
u
g
2
9
64
3
16
    =
u
g
2
21
8
Option (d) is correct.
7. Let the projectile hits the inclined plane at 
P at time t
PQ u gt = - ( sin ) a
1
2
2
Further, QR vt u t = - ( cos ) a
and tan b =
PQ
QR
i.e., OQ PQ tan b =
t v u gt t [ cos ] tan sin - = -
é
ë
ê
ù
û
ú
a b a
1
2
or      10 1 60 30 [ cos ] tan - ° °
      = ° - ´ ´
é
ë
ê
ù
û
ú
10 60
1
2
10 sin t
or          
5
3
5 3 5 = - t
or           t = - 3
1
3
                    =
2
3
 s
8. y y x
2
2 2 + + =
\ 2 2 y
dy
dt
dy
dt
dx
dt
+ =
or ( ) 2 2 y
dy
dt
dx
dt
+ =
or
dx
dt
y = + 10 1 ( )
or
d x
dt
dy
dt
2
2
10 =
æ
è
ç
ö
ø
÷
= 50 m/s
2
Acceleration of the particle
       =
d x
dt
2
2
       =50 m/s
2
[as 
dy
dt
= constant]
9. tan tan tan a q = + f
              = + tan tan q q = 2 tan q
\ q
a
=
æ
è
ç
ö
ø
÷
-
tan
tan
1
2
Option (c) is correct.
10. At any time t,
Horizontal distance between particles
x t = - + 20 3 10 3 10 3 ( )
= - 20 3 1 ( ) t
Vertical distance between particles at
time t
y t t = - = ( ) 30 10 20
Distance between particles at time t
        D x y = +
2 2
or D t t
2 2 2
20 3 1 20 = - + [ ( )] [ ]
      = - + 400 3 1 400
2 2
[ ( ) ] t t
For D to minimum
dD
dt
= 0
3 2 1 1 2 0 ´ - - + = ( )( ) t t
or 6 6 2 0 t t - + =
i.e., t =
3
4
 s
\ D
min
2
2 2
400 3 1
3
4
400
3
4
= -
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
+
æ
è
ç
ö
ø
÷
= + ´ 75 75 3
Þ D
min
= 10 3 m
Option (b) is correct.
68 | Mechanics-1
20Ö3m
20Ö3 m/s
20 m/s
30° 60°
q f
H
R/2 R/2
a
b
v
v = 10 Ö3 m/s
b
P (time = t)
R
Q
(u cos a) t
vt
u
R
a
u = 10 Ö3 m/s
i.e., v = u
11. Let time of flight = T
As, horizontal component of the velocity of 
the particle will not change
u v cos ( ) cos ( ) 90 90 ° - = ° - a b
or u v sin sin a b = …(i)
Using v u at = + for the horizontal
component of velocity
( cos ) ( cos ) ( ) - = + + - v u g T b a
Þ gT u v = + cos cos a b
or T
u v
g
=
+ cos cos a b
…(ii)
Now, as a b = = ° ( ) 30 , u v = from Eq. (i)
Substituting v u = and a b = = ° 30 in Eq.
(ii)
T
u u
g
=
° + ° cos cos 30 30
=
° 2 30 u
g
cos
=
u
g
3
 
Option (b) is correct.
12. PQ
u
g
=
2
2
sin cos q q
or 2
2 90 90
2
a
u
g
cos
sin ( )cos ( )
a
a a
=
° - ° -
or a
u
g
=
2
sin a
or u ag
2
2 =         (as a = ° 30 )
or     u
2
2 = ´ ´ 4.9 9.8
or u = 9.8 m/s
Option (a) is correct.
More than One Correct Options
1. Two particles projected at angles a and b
with same speed ( ) =m will have same
range if
a b + = ° 90
Option (a) is correct.
R
u
g
=
2
2
sin cos a a
h
1
 (maximum height attained by first)
=
u
g
2 2
2
sin a
h
2
 (maximum height attained by second)
=
u
g
2 2
2
sin b
\      h h
u
g
1 2
2
2
=
sin sin a b
  =
° - u
g
2
90
2
sin sin ( ) a a
=
u
g
2
2
sin cos a a
 =
R
4
Option (b) is correct.
t
1
 (time of flight of first) =
2u
g
sin a
t
2
 (time of flight of second) =
2u
g
sin b
\ 
t
t
1
2
90
= =
° -
=
sin
sin
sin
sin ( )
tan
a
b
a
a
a
Option (c) is correct.
Also,
h
h
1
2
= =
sin
sin
tan
a
b
a
Option (d) is correct.
2. Horizontal displacement in time t (Time of 
flight)
x at =
1
2
2
…(i)
Projectile Motion 69
Q
u
a
P
b
u
a b 
q
b
a
a = constant
Path of particle
Q
R
P
a = g
Wind
Page 4


2. t
d
u
= also h gt =
1
2
2
\ h g
d
u
= ×
1
2
2
2
Þ d
u h
g
2
2
2
=
Option (b) is correct.
3. R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
Substituting v = 10 m/s
           g =10 m/s
2
         a b + = ° 90
and        b = ° 30
R =
°
° + ° + °
( )
cos
[sin( ) sin ]
10
10 30
90 60 30
2
2
   =
æ
è
ç
ö
ø
÷
+
é
ë
ê
ù
û
ú
10
3
4
1
2
1
2
 =
40
3
 m 
Option (c) is correct.
4. y x g
x
u
= - tan
sin cos
q
q q
1
2
2
2
dy
dx
x
u
= - tan
sin cos
q
q q
1
2
2
2
Slope of trajectory = - tan
sin cos
q
q q
x
u
2
Substituting x u t = ( cos ) q
    
dy
dx
u t
u
= - tan
( cos )
sin cos
q
q
q q
2
  = - tan
sin
q
q
t
u
Option (a) is correct.
5.         y x =b
2
\        
dy
dt
x
dx
dt
= × b 2
and
d y
dt
x
d x
dt
dx
dt
2
2
2
2
2
2 = × +
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
b
or  a b = +
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
2
2
2
2
x
d x
dt
dx
dt
Now, as 
d x
dt
2
2
0 = 
(acceleration being along y-axis only)
dx
dt
=
a
b 2
Option (d) is correct.
6. As the projectile hits the inclined plane
horizontally
PQ H
u
g
= =
max
sin
2 2
2
a
and OQ
u
g
=
2
sin cos a a
\ Range on inclined plane
  R PQ OQ ¢ = + ( ) ( )
2 2
    =
æ
è
ç
ö
ø
÷
+
æ
è
ç
ö
ø
÷
u
g
u
g
2 2
2
2
2
2
sin sin cos a a a
    = +
u
g
2 4
2 2
4
sin
sin cos
a
a a
    =
æ
è
ç
ö
ø
÷
+
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷
u
g
2
4 2
2
1
4
3
4
3
2
1
2
Projectile Motion 67
Q
P
R'
O
u
a
b
x
Slope
t
Slope
T
tan q
O
T/2
T = Time of flight
h
u
d
time = t
    = +
u
g
2
9
64
3
16
    =
u
g
2
21
8
Option (d) is correct.
7. Let the projectile hits the inclined plane at 
P at time t
PQ u gt = - ( sin ) a
1
2
2
Further, QR vt u t = - ( cos ) a
and tan b =
PQ
QR
i.e., OQ PQ tan b =
t v u gt t [ cos ] tan sin - = -
é
ë
ê
ù
û
ú
a b a
1
2
or      10 1 60 30 [ cos ] tan - ° °
      = ° - ´ ´
é
ë
ê
ù
û
ú
10 60
1
2
10 sin t
or          
5
3
5 3 5 = - t
or           t = - 3
1
3
                    =
2
3
 s
8. y y x
2
2 2 + + =
\ 2 2 y
dy
dt
dy
dt
dx
dt
+ =
or ( ) 2 2 y
dy
dt
dx
dt
+ =
or
dx
dt
y = + 10 1 ( )
or
d x
dt
dy
dt
2
2
10 =
æ
è
ç
ö
ø
÷
= 50 m/s
2
Acceleration of the particle
       =
d x
dt
2
2
       =50 m/s
2
[as 
dy
dt
= constant]
9. tan tan tan a q = + f
              = + tan tan q q = 2 tan q
\ q
a
=
æ
è
ç
ö
ø
÷
-
tan
tan
1
2
Option (c) is correct.
10. At any time t,
Horizontal distance between particles
x t = - + 20 3 10 3 10 3 ( )
= - 20 3 1 ( ) t
Vertical distance between particles at
time t
y t t = - = ( ) 30 10 20
Distance between particles at time t
        D x y = +
2 2
or D t t
2 2 2
20 3 1 20 = - + [ ( )] [ ]
      = - + 400 3 1 400
2 2
[ ( ) ] t t
For D to minimum
dD
dt
= 0
3 2 1 1 2 0 ´ - - + = ( )( ) t t
or 6 6 2 0 t t - + =
i.e., t =
3
4
 s
\ D
min
2
2 2
400 3 1
3
4
400
3
4
= -
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
+
æ
è
ç
ö
ø
÷
= + ´ 75 75 3
Þ D
min
= 10 3 m
Option (b) is correct.
68 | Mechanics-1
20Ö3m
20Ö3 m/s
20 m/s
30° 60°
q f
H
R/2 R/2
a
b
v
v = 10 Ö3 m/s
b
P (time = t)
R
Q
(u cos a) t
vt
u
R
a
u = 10 Ö3 m/s
i.e., v = u
11. Let time of flight = T
As, horizontal component of the velocity of 
the particle will not change
u v cos ( ) cos ( ) 90 90 ° - = ° - a b
or u v sin sin a b = …(i)
Using v u at = + for the horizontal
component of velocity
( cos ) ( cos ) ( ) - = + + - v u g T b a
Þ gT u v = + cos cos a b
or T
u v
g
=
+ cos cos a b
…(ii)
Now, as a b = = ° ( ) 30 , u v = from Eq. (i)
Substituting v u = and a b = = ° 30 in Eq.
(ii)
T
u u
g
=
° + ° cos cos 30 30
=
° 2 30 u
g
cos
=
u
g
3
 
Option (b) is correct.
12. PQ
u
g
=
2
2
sin cos q q
or 2
2 90 90
2
a
u
g
cos
sin ( )cos ( )
a
a a
=
° - ° -
or a
u
g
=
2
sin a
or u ag
2
2 =         (as a = ° 30 )
or     u
2
2 = ´ ´ 4.9 9.8
or u = 9.8 m/s
Option (a) is correct.
More than One Correct Options
1. Two particles projected at angles a and b
with same speed ( ) =m will have same
range if
a b + = ° 90
Option (a) is correct.
R
u
g
=
2
2
sin cos a a
h
1
 (maximum height attained by first)
=
u
g
2 2
2
sin a
h
2
 (maximum height attained by second)
=
u
g
2 2
2
sin b
\      h h
u
g
1 2
2
2
=
sin sin a b
  =
° - u
g
2
90
2
sin sin ( ) a a
=
u
g
2
2
sin cos a a
 =
R
4
Option (b) is correct.
t
1
 (time of flight of first) =
2u
g
sin a
t
2
 (time of flight of second) =
2u
g
sin b
\ 
t
t
1
2
90
= =
° -
=
sin
sin
sin
sin ( )
tan
a
b
a
a
a
Option (c) is correct.
Also,
h
h
1
2
= =
sin
sin
tan
a
b
a
Option (d) is correct.
2. Horizontal displacement in time t (Time of 
flight)
x at =
1
2
2
…(i)
Projectile Motion 69
Q
u
a
P
b
u
a b 
q
b
a
a = constant
Path of particle
Q
R
P
a = g
Wind
Vertical displacement in time t
y gt =
1
2
2
…(ii)
\ 
y
x
g
a
=
or y = constant x [as a is constant]
Option (a) is correct.
Substituting y = 49 m in Eqs. (ii)
49
1
2
2
= ´ ´ 9.8 t
Þ t = = 10 3.16 s
PQ PR > = ( ) 40 m
Option (d) is correct.
3. As there will be no change in the
horizontal component of the velocity of the 
particle
w v cos cos f = q
or w v cos ( ) cos 90° - = q q
or w v sin cos q q =
or w v = cos q
Option (b) is correct.
Option (a) is incorrect.
Vertical velocity of particle at Q
            = f wsin
            = ° - wsin( ) 90 q
            = wcosq
Using, v u at = +
( cos ) ( sin ) ( ) - = + + - w v g T q q
[where T = time from P to Q]
Þ gT v w = + sin cos q q
      = + v v sin cot cos q q q
    = + × v v sin
cos
sin
q
q
q
2
=vcosecq   
\ t
v
g
=
cosecq
    
Option (c) is correct.
Option (d) is incorrect.
4. v i j
®
= + 10 10
^ ^
If q be the angle of projection 
tan q = = =
v
v
y
x
10
10
1
i.e., q = ° 45      
Option (a) correct.
Using relation, s ut at = +
1
2
2
( ) ( ) - = + - 15 10
1
2
10
2
t t
i.e., t t
2
2 3 0 - - =
or ( ) ( ) t t - + = 3 1 0
\ t = 3 s
Option (b) is incorrect.
Horizontal range of particle
= × v t
x
       = ´ = 10 3 30 m
Option (c) is incorrect.
Maximum height of projectile from ground
            = + H 15
= +
u
g
2 2
2
15
sin q
            = +
v
g
y
2
2
15
            =
´
+
( ) 10
2 10
15
2
            =20 m
Option (d) is correct.
5. Average velocity between any two points
can’t remain constant as in projectile
motion velocity changes both in
magnitude and direction.
Option (a) is incorrect.
Similarly option (b) is also incorrect.
70 | Mechanics-1
– 15m
v
2
a = – 10 m/s
H
t
O
Q
w cos f
w
f
v
P
q
q + f = 90°
v cos q
Page 5


2. t
d
u
= also h gt =
1
2
2
\ h g
d
u
= ×
1
2
2
2
Þ d
u h
g
2
2
2
=
Option (b) is correct.
3. R
u
g
= + +
2
2
2
cos
[sin ( ) sin ]
b
a b b
Substituting v = 10 m/s
           g =10 m/s
2
         a b + = ° 90
and        b = ° 30
R =
°
° + ° + °
( )
cos
[sin( ) sin ]
10
10 30
90 60 30
2
2
   =
æ
è
ç
ö
ø
÷
+
é
ë
ê
ù
û
ú
10
3
4
1
2
1
2
 =
40
3
 m 
Option (c) is correct.
4. y x g
x
u
= - tan
sin cos
q
q q
1
2
2
2
dy
dx
x
u
= - tan
sin cos
q
q q
1
2
2
2
Slope of trajectory = - tan
sin cos
q
q q
x
u
2
Substituting x u t = ( cos ) q
    
dy
dx
u t
u
= - tan
( cos )
sin cos
q
q
q q
2
  = - tan
sin
q
q
t
u
Option (a) is correct.
5.         y x =b
2
\        
dy
dt
x
dx
dt
= × b 2
and
d y
dt
x
d x
dt
dx
dt
2
2
2
2
2
2 = × +
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
b
or  a b = +
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
2
2
2
2
x
d x
dt
dx
dt
Now, as 
d x
dt
2
2
0 = 
(acceleration being along y-axis only)
dx
dt
=
a
b 2
Option (d) is correct.
6. As the projectile hits the inclined plane
horizontally
PQ H
u
g
= =
max
sin
2 2
2
a
and OQ
u
g
=
2
sin cos a a
\ Range on inclined plane
  R PQ OQ ¢ = + ( ) ( )
2 2
    =
æ
è
ç
ö
ø
÷
+
æ
è
ç
ö
ø
÷
u
g
u
g
2 2
2
2
2
2
sin sin cos a a a
    = +
u
g
2 4
2 2
4
sin
sin cos
a
a a
    =
æ
è
ç
ö
ø
÷
+
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷
u
g
2
4 2
2
1
4
3
4
3
2
1
2
Projectile Motion 67
Q
P
R'
O
u
a
b
x
Slope
t
Slope
T
tan q
O
T/2
T = Time of flight
h
u
d
time = t
    = +
u
g
2
9
64
3
16
    =
u
g
2
21
8
Option (d) is correct.
7. Let the projectile hits the inclined plane at 
P at time t
PQ u gt = - ( sin ) a
1
2
2
Further, QR vt u t = - ( cos ) a
and tan b =
PQ
QR
i.e., OQ PQ tan b =
t v u gt t [ cos ] tan sin - = -
é
ë
ê
ù
û
ú
a b a
1
2
or      10 1 60 30 [ cos ] tan - ° °
      = ° - ´ ´
é
ë
ê
ù
û
ú
10 60
1
2
10 sin t
or          
5
3
5 3 5 = - t
or           t = - 3
1
3
                    =
2
3
 s
8. y y x
2
2 2 + + =
\ 2 2 y
dy
dt
dy
dt
dx
dt
+ =
or ( ) 2 2 y
dy
dt
dx
dt
+ =
or
dx
dt
y = + 10 1 ( )
or
d x
dt
dy
dt
2
2
10 =
æ
è
ç
ö
ø
÷
= 50 m/s
2
Acceleration of the particle
       =
d x
dt
2
2
       =50 m/s
2
[as 
dy
dt
= constant]
9. tan tan tan a q = + f
              = + tan tan q q = 2 tan q
\ q
a
=
æ
è
ç
ö
ø
÷
-
tan
tan
1
2
Option (c) is correct.
10. At any time t,
Horizontal distance between particles
x t = - + 20 3 10 3 10 3 ( )
= - 20 3 1 ( ) t
Vertical distance between particles at
time t
y t t = - = ( ) 30 10 20
Distance between particles at time t
        D x y = +
2 2
or D t t
2 2 2
20 3 1 20 = - + [ ( )] [ ]
      = - + 400 3 1 400
2 2
[ ( ) ] t t
For D to minimum
dD
dt
= 0
3 2 1 1 2 0 ´ - - + = ( )( ) t t
or 6 6 2 0 t t - + =
i.e., t =
3
4
 s
\ D
min
2
2 2
400 3 1
3
4
400
3
4
= -
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
+
æ
è
ç
ö
ø
÷
= + ´ 75 75 3
Þ D
min
= 10 3 m
Option (b) is correct.
68 | Mechanics-1
20Ö3m
20Ö3 m/s
20 m/s
30° 60°
q f
H
R/2 R/2
a
b
v
v = 10 Ö3 m/s
b
P (time = t)
R
Q
(u cos a) t
vt
u
R
a
u = 10 Ö3 m/s
i.e., v = u
11. Let time of flight = T
As, horizontal component of the velocity of 
the particle will not change
u v cos ( ) cos ( ) 90 90 ° - = ° - a b
or u v sin sin a b = …(i)
Using v u at = + for the horizontal
component of velocity
( cos ) ( cos ) ( ) - = + + - v u g T b a
Þ gT u v = + cos cos a b
or T
u v
g
=
+ cos cos a b
…(ii)
Now, as a b = = ° ( ) 30 , u v = from Eq. (i)
Substituting v u = and a b = = ° 30 in Eq.
(ii)
T
u u
g
=
° + ° cos cos 30 30
=
° 2 30 u
g
cos
=
u
g
3
 
Option (b) is correct.
12. PQ
u
g
=
2
2
sin cos q q
or 2
2 90 90
2
a
u
g
cos
sin ( )cos ( )
a
a a
=
° - ° -
or a
u
g
=
2
sin a
or u ag
2
2 =         (as a = ° 30 )
or     u
2
2 = ´ ´ 4.9 9.8
or u = 9.8 m/s
Option (a) is correct.
More than One Correct Options
1. Two particles projected at angles a and b
with same speed ( ) =m will have same
range if
a b + = ° 90
Option (a) is correct.
R
u
g
=
2
2
sin cos a a
h
1
 (maximum height attained by first)
=
u
g
2 2
2
sin a
h
2
 (maximum height attained by second)
=
u
g
2 2
2
sin b
\      h h
u
g
1 2
2
2
=
sin sin a b
  =
° - u
g
2
90
2
sin sin ( ) a a
=
u
g
2
2
sin cos a a
 =
R
4
Option (b) is correct.
t
1
 (time of flight of first) =
2u
g
sin a
t
2
 (time of flight of second) =
2u
g
sin b
\ 
t
t
1
2
90
= =
° -
=
sin
sin
sin
sin ( )
tan
a
b
a
a
a
Option (c) is correct.
Also,
h
h
1
2
= =
sin
sin
tan
a
b
a
Option (d) is correct.
2. Horizontal displacement in time t (Time of 
flight)
x at =
1
2
2
…(i)
Projectile Motion 69
Q
u
a
P
b
u
a b 
q
b
a
a = constant
Path of particle
Q
R
P
a = g
Wind
Vertical displacement in time t
y gt =
1
2
2
…(ii)
\ 
y
x
g
a
=
or y = constant x [as a is constant]
Option (a) is correct.
Substituting y = 49 m in Eqs. (ii)
49
1
2
2
= ´ ´ 9.8 t
Þ t = = 10 3.16 s
PQ PR > = ( ) 40 m
Option (d) is correct.
3. As there will be no change in the
horizontal component of the velocity of the 
particle
w v cos cos f = q
or w v cos ( ) cos 90° - = q q
or w v sin cos q q =
or w v = cos q
Option (b) is correct.
Option (a) is incorrect.
Vertical velocity of particle at Q
            = f wsin
            = ° - wsin( ) 90 q
            = wcosq
Using, v u at = +
( cos ) ( sin ) ( ) - = + + - w v g T q q
[where T = time from P to Q]
Þ gT v w = + sin cos q q
      = + v v sin cot cos q q q
    = + × v v sin
cos
sin
q
q
q
2
=vcosecq   
\ t
v
g
=
cosecq
    
Option (c) is correct.
Option (d) is incorrect.
4. v i j
®
= + 10 10
^ ^
If q be the angle of projection 
tan q = = =
v
v
y
x
10
10
1
i.e., q = ° 45      
Option (a) correct.
Using relation, s ut at = +
1
2
2
( ) ( ) - = + - 15 10
1
2
10
2
t t
i.e., t t
2
2 3 0 - - =
or ( ) ( ) t t - + = 3 1 0
\ t = 3 s
Option (b) is incorrect.
Horizontal range of particle
= × v t
x
       = ´ = 10 3 30 m
Option (c) is incorrect.
Maximum height of projectile from ground
            = + H 15
= +
u
g
2 2
2
15
sin q
            = +
v
g
y
2
2
15
            =
´
+
( ) 10
2 10
15
2
            =20 m
Option (d) is correct.
5. Average velocity between any two points
can’t remain constant as in projectile
motion velocity changes both in
magnitude and direction.
Option (a) is incorrect.
Similarly option (b) is also incorrect.
70 | Mechanics-1
– 15m
v
2
a = – 10 m/s
H
t
O
Q
w cos f
w
f
v
P
q
q + f = 90°
v cos q
In projectile motion
a
®
= constant
\
d
dt
v
®
= constant 
Option (c) is correct.
also
d
dt
2
2
0
v
®
=
Option (d) is correct.
6. ( ) ( sin ) ( ) + = + + - h u t g t a
1
2
2
Þ gt u t h
2
2 2 0 - + = ( sin ) a …(i)
t t
u
g
1 2
2
+ =
sin a
, t t
h
g
1 2
2
=
( ) ( ) t t t t t t
2 1
2
2 1
2
1 2
4 - = + -
= = ×
4
4
2
2 2
u
g
h
g
sin a
\  t
H h
g
AB
=
- 8 ( )
[where H =    Maximum height]
Þ 2
8 15
10
=
- ( ) H
\ H = 20 m
Option (b) is correct.
i.e.,
u
g
2 2
2
20
sin a
=
u g sin a = 40
          = 20 m/s
Option (c) is correct.
    ( cos ) u t a
AB
= 40
Þ ucosa =
40
2
                     = 20 m/s  
Option (d) is correct.
Substituting values of g, usina and h in
Eq. (i)
10 40 30 0
2
t t - + =
i.e., t t
2
4 3 0 - + =
\ t = 1 or 3
t = 1 s for A and t = 3 s for B.
Option  (a) is correct.
Match the Columns
1. (a) At t = 2 s
horizontal distance between  1 and 2
= horizontal displacement of 2
    = ´ 10 1 =10 m 
\ (a) ® (p)
(b) Vertical distance between the two  
at t = 2 s = 30 m
\ (b) ® (s)
(c) Relative horizontal component of
velocity
           =10 -0 =10 m/s
\ (c) ® (p)
(d) Relative vertical component of velocity
(at t = 2 s)
= Velocity of 1 at t = 2 s
- Vertical velocity of 2 at t = 2 s
= Velocity of 1 at t = 2 s
- Velocity of 1 at t = 1 s
         = - ( ) ( ) 2 1 g g = g
         =10 m/s
(d) ® (p).
Projectile Motion 71
a
40 m
B
t = t
1
t = t
2
h 15 m
20 m
v
A
10 m S
1
S
2
S
3
30 m
50 m
1 (at t = 2 s)
2 (t = 1 s)
at t = 2 s
2
1 (at t = 1s)
10 m/s
S : S : S
123
= 1 : 3 : 5
Read More
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Complete Syllabus of NEET

Dynamic Test

Content Category

Related Searches

Free

,

MCQs

,

Solution by D C Pandey

,

past year papers

,

Sample Paper

,

Objective type Questions

,

mock tests for examination

,

Viva Questions

,

shortcuts and tricks

,

Chapter 4 - Projectile Motion (Part - 3) - Physics

,

Chapter 4 - Projectile Motion (Part - 3) - Physics

,

ppt

,

Previous Year Questions with Solutions

,

Solution by D C Pandey

,

pdf

,

Semester Notes

,

NEET NEET Notes | EduRev

,

Chapter 4 - Projectile Motion (Part - 3) - Physics

,

video lectures

,

practice quizzes

,

Summary

,

Important questions

,

NEET NEET Notes | EduRev

,

Solution by D C Pandey

,

Extra Questions

,

NEET NEET Notes | EduRev

,

Exam

,

study material

;