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

## DC Pandey (Questions & Solutions) of Physics: NEET

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## 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 = â€¦(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
â€¦(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
â€¦(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 = â€¦(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
â€¦(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
â€¦(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
â€¦(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
canâ€™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
â€“ 15m
v
2
a = â€“ 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 = â€¦(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
â€¦(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
â€¦(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
â€¦(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
canâ€™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
â€“ 15m
v
2
a = â€“ 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 â€¦(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
```
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