NEET Exam > NEET Notes > DC Pandey Solutions for NEET Physics > DC Pandey Solutions: Projectile Motion- 3
DC Pandey Solutions: Projectile Motion- 3 | DC Pandey Solutions for NEET Physics PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
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.
About this Document
|
4.93/5 Rating |
|
Nov 24, 2024 Last updated |
Document Description: DC Pandey Solutions: Projectile Motion- 3 for NEET 2024 is part of DC Pandey Solutions for NEET Physics preparation.
The notes and questions for DC Pandey Solutions: Projectile Motion- 3 have been prepared according to the NEET exam syllabus. Information about DC Pandey Solutions: Projectile Motion- 3 covers topics
like and DC Pandey Solutions: Projectile Motion- 3 Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for DC Pandey Solutions: Projectile Motion- 3.
Introduction of DC Pandey Solutions: Projectile Motion- 3 in English is available as part of our DC Pandey Solutions for NEET Physics
for NEET & DC Pandey Solutions: Projectile Motion- 3 in Hindi for DC Pandey Solutions for NEET Physics course.
Download more important topics related with notes, lectures and mock test series for NEET
Exam by signing up for free. NEET: DC Pandey Solutions: Projectile Motion- 3 | DC Pandey Solutions for NEET Physics
Description
Full syllabus notes, lecture & questions for DC Pandey Solutions: Projectile Motion- 3 | DC Pandey Solutions for NEET Physics - NEET | Plus excerises question with solution to help you revise complete syllabus for DC Pandey Solutions for NEET Physics | Best notes, free PDF download
Information about DC Pandey Solutions: Projectile Motion- 3
In this doc you can find the meaning of DC Pandey Solutions: Projectile Motion- 3 defined & explained in the simplest way possible. Besides explaining types of
DC Pandey Solutions: Projectile Motion- 3 theory, EduRev gives you an ample number of questions to practice DC Pandey Solutions: Projectile Motion- 3 tests, examples and also practice NEET
tests
|
Explore Courses for NEET exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Important questions
,DC Pandey Solutions: Projectile Motion- 3 | DC Pandey Solutions for NEET Physics
,past year papers
,practice quizzes
,mock tests for examination
,MCQs
,shortcuts and tricks
,study material
,Viva Questions
,Semester Notes
,Objective type Questions
,Summary
,video lectures
,Extra Questions
,Free
,ppt
,Sample Paper
,DC Pandey Solutions: Projectile Motion- 3 | DC Pandey Solutions for NEET Physics
,Exam
,DC Pandey Solutions: Projectile Motion- 3 | DC Pandey Solutions for NEET Physics
,Previous Year Questions with Solutions
;
Additional Information about DC Pandey Solutions: Projectile Motion- 3 for NEET Preparation
DC Pandey Solutions: Projectile Motion- 3 Free PDF Download
The DC Pandey Solutions: Projectile Motion- 3 is an invaluable resource that delves deep into the core of the NEET exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the DC Pandey Solutions: Projectile Motion- 3 now and kickstart your journey towards success in the NEET exam.
Importance of DC Pandey Solutions: Projectile Motion- 3
The importance of DC Pandey Solutions: Projectile Motion- 3 cannot be overstated, especially for NEET aspirants.
This document holds the key to success in the NEET exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
DC Pandey Solutions: Projectile Motion- 3 Notes
DC Pandey Solutions: Projectile Motion- 3 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to DC Pandey Solutions: Projectile Motion- 3.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, DC Pandey Solutions: Projectile Motion- 3 Notes on EduRev are your ultimate resource for success.
DC Pandey Solutions: Projectile Motion- 3 NEET Questions
The "DC Pandey Solutions: Projectile Motion- 3 NEET Questions" guide is a valuable resource for all aspiring students preparing for the
NEET exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study DC Pandey Solutions: Projectile Motion- 3 on the App
Students of NEET can study DC Pandey Solutions: Projectile Motion- 3 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the DC Pandey Solutions: Projectile Motion- 3,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of DC Pandey Solutions: Projectile Motion- 3 is prepared as per the latest NEET syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup