NEET Exam > NEET Notes > DC Pandey Solutions for NEET Physics > DC Pandey Solutions: Projectile Motion- 3
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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
FAQs on DC Pandey Solutions: Projectile Motion- 3 - DC Pandey Solutions for NEET Physics
| 1. What is projectile motion? |  |
Ans. Projectile motion is a type of motion where an object is thrown or projected into the air, and it follows a curved path due to gravity. The motion of an object in projectile motion can be described by its initial velocity, angle of projection, and the force of gravity acting on it.
| 2. How do you calculate the maximum height of a projectile? | |
Ans. To calculate the maximum height of a projectile, we can use the formula Hmax = (u^2 * sin^2θ) / 2g, where u is the initial velocity of the projectile, θ is the angle of projection, and g is the acceleration due to gravity.
| 3. What is the range of a projectile? | |
Ans. The range of a projectile is the horizontal distance it travels before hitting the ground. It can be calculated using the formula R = (u^2 * sin2θ) / g, where u is the initial velocity of the projectile, θ is the angle of projection, and g is the acceleration due to gravity.
| 4. What happens if the angle of projection is 90 degrees? | |
Ans. If the angle of projection is 90 degrees, the projectile will travel in a purely vertical direction, and its range will be zero. The maximum height of the projectile can be calculated using the formula Hmax = (u^2) / 2g, where u is the initial velocity of the projectile, and g is the acceleration due to gravity.
| 5. How can projectile motion be applied in real life? | |
Ans. Projectile motion has many real-life applications, such as in sports like basketball, football, and baseball. It is also used in military applications, such as calculating the trajectory of missiles and bombs. Additionally, it is used in engineering design, such as in the design of roller coasters and amusement park rides.
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Nov 21, 2024 Last updated |
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The content of DC Pandey Solutions: Projectile Motion- 3 is prepared as per the latest NEET syllabus.
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