Chapter 4 - Projectile Motion (Part - 1) - 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 - 1) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

 Page 1


Introductory Exercise 4.1
1. A particle projected at any angle with
horizontal will always move in a plane
and thus projectile motion is a
2-dimensional motion. The statement is
thus false.
2. At high speed the projectile may go to a
place where acceleration due to gravity
has some different value and as such the
motion may not be uniform accelerated.
The statement is thus true.
3. See article 4.1.
4. u = 40 2 m/s, q = ° 45
As horizontal acceleration would be zero.
v u u
x x
= = = cos q 40 m/s
 s u t u t
x x
= = = ( cos ) q 80 m
A : position of particle at time = 0.
B : position of particle at time = t.
As vertical acceleration would be - g
          v u gt
y y
= -
= = u gt sinq
             = - 40 20
             =20 m/s
s u t gt
y y
= × -
1
2
2
  
= - ´ ´ = 80
1
2
10 2 60
2
 m
\ v v v
x y
= +
2 2
= + 40 20
2 2
= 20 5 m/s
tan f = =
v
v
y
x
20
40
i.e., f =
æ
è
ç
ö
ø
÷
-
tan
1
1
2
s s s
x y
= +
2 2
= + 80 60
2 2
= 100 m
tan a =
s
s
y
x
=
60
80
i.e., a =
æ
è
ç
ö
ø
÷
-
tan
1
3
4
5. s u t g t
y y
= + -
1
2
2
( )
or s u t gt
y
= - ( sin ) q
1
2
2
or 15 20
1
2
2
= - t gt
or t t
2
4 3 0 = + =
i.e., t = 1  s and 3 s
6. See figure to the answer to question no. 4.
u = 40 m/s, q = ° 60
\ u
x
= ° = 40 60 20 cos m/s
Projectile Motion
4
s
y
g
f
v
v
y
v
x
u
x
s
x
A
a
u
u
y
B
Page 2


Introductory Exercise 4.1
1. A particle projected at any angle with
horizontal will always move in a plane
and thus projectile motion is a
2-dimensional motion. The statement is
thus false.
2. At high speed the projectile may go to a
place where acceleration due to gravity
has some different value and as such the
motion may not be uniform accelerated.
The statement is thus true.
3. See article 4.1.
4. u = 40 2 m/s, q = ° 45
As horizontal acceleration would be zero.
v u u
x x
= = = cos q 40 m/s
 s u t u t
x x
= = = ( cos ) q 80 m
A : position of particle at time = 0.
B : position of particle at time = t.
As vertical acceleration would be - g
          v u gt
y y
= -
= = u gt sinq
             = - 40 20
             =20 m/s
s u t gt
y y
= × -
1
2
2
  
= - ´ ´ = 80
1
2
10 2 60
2
 m
\ v v v
x y
= +
2 2
= + 40 20
2 2
= 20 5 m/s
tan f = =
v
v
y
x
20
40
i.e., f =
æ
è
ç
ö
ø
÷
-
tan
1
1
2
s s s
x y
= +
2 2
= + 80 60
2 2
= 100 m
tan a =
s
s
y
x
=
60
80
i.e., a =
æ
è
ç
ö
ø
÷
-
tan
1
3
4
5. s u t g t
y y
= + -
1
2
2
( )
or s u t gt
y
= - ( sin ) q
1
2
2
or 15 20
1
2
2
= - t gt
or t t
2
4 3 0 = + =
i.e., t = 1  s and 3 s
6. See figure to the answer to question no. 4.
u = 40 m/s, q = ° 60
\ u
x
= ° = 40 60 20 cos m/s
Projectile Motion
4
s
y
g
f
v
v
y
v
x
u
x
s
x
A
a
u
u
y
B
Thus, v u
x x
= = 20 m/s
As f = ° 45
tan f = =
v
v
y
x
1
\ y v
y x
= = 20 m/s
Thus, v v v v
x y x
= + =
2 2
2
= 20 2 m/s
Before reaching highest point
v u g t
y y
= + - ( )
\ 20 40 60 10 = ° - sin t
or 2 2 3 = - t
Þ t = - 2 3 1 ( ) s
After attaining highest point
- = - 20 40 3 10t
i.e., - = - 2 2 3 t
or t = + 2 3 1 ( ) s
7. Average velocity =
Range
Time of flight
=
u g
u g
2
2
sin 2a
a /
/
sin
= ´
2
2
2
u
g
g
u
sin cos
sin
a a
a
= ucos a
8. Change in velocity
          = - - + ( sin ) ( sin ) u u a a
          = -2usina
          =2usina (downward)
9. Formulae for R T , and H
max
 will be same
if the projection point and the point where
the particle lands are same and lie on a
horizontal line.
10. r i j
®
= + - [ ( ) ]
^ ^
3 4 5
2
t t t m
y-coordinate will be zero when
4 5 0
2
t t - =
i.e., t = 0 s, 
4
5
 s
t = 0 belongs to the initial point of
projection of the particle.
\ r i
®
= 0
^
 m
i.e., x = 0 m
At t = 0.8 s,
r i
®
= 2.4
^
 m
i.e., x = 2.4 m
11. v u
x x
= = 10 m/s
      =10 m/s
        v u
x x
= = 10 m/s
        v v
x
= f cos
Þ    cosf=
v
v
x
          =
10
v
          =
10
10
[as, v =
20
2
 (given)]
          =1
i.e.,       f = ° 0
\ Speed will be half of its initial value at
the highest point where f = ° 0 .
Thus,       t
u
g
=
sin q
 =
°
=
20 60
10
3
sin
 s
52 | Mechanics-1
a
a
u – u sin a
u sin a
t = T
u cos a
+ u sin a u
60°
u = 20 m/s
v
x
v
f
u = 20 cos 60°
x
Page 3


Introductory Exercise 4.1
1. A particle projected at any angle with
horizontal will always move in a plane
and thus projectile motion is a
2-dimensional motion. The statement is
thus false.
2. At high speed the projectile may go to a
place where acceleration due to gravity
has some different value and as such the
motion may not be uniform accelerated.
The statement is thus true.
3. See article 4.1.
4. u = 40 2 m/s, q = ° 45
As horizontal acceleration would be zero.
v u u
x x
= = = cos q 40 m/s
 s u t u t
x x
= = = ( cos ) q 80 m
A : position of particle at time = 0.
B : position of particle at time = t.
As vertical acceleration would be - g
          v u gt
y y
= -
= = u gt sinq
             = - 40 20
             =20 m/s
s u t gt
y y
= × -
1
2
2
  
= - ´ ´ = 80
1
2
10 2 60
2
 m
\ v v v
x y
= +
2 2
= + 40 20
2 2
= 20 5 m/s
tan f = =
v
v
y
x
20
40
i.e., f =
æ
è
ç
ö
ø
÷
-
tan
1
1
2
s s s
x y
= +
2 2
= + 80 60
2 2
= 100 m
tan a =
s
s
y
x
=
60
80
i.e., a =
æ
è
ç
ö
ø
÷
-
tan
1
3
4
5. s u t g t
y y
= + -
1
2
2
( )
or s u t gt
y
= - ( sin ) q
1
2
2
or 15 20
1
2
2
= - t gt
or t t
2
4 3 0 = + =
i.e., t = 1  s and 3 s
6. See figure to the answer to question no. 4.
u = 40 m/s, q = ° 60
\ u
x
= ° = 40 60 20 cos m/s
Projectile Motion
4
s
y
g
f
v
v
y
v
x
u
x
s
x
A
a
u
u
y
B
Thus, v u
x x
= = 20 m/s
As f = ° 45
tan f = =
v
v
y
x
1
\ y v
y x
= = 20 m/s
Thus, v v v v
x y x
= + =
2 2
2
= 20 2 m/s
Before reaching highest point
v u g t
y y
= + - ( )
\ 20 40 60 10 = ° - sin t
or 2 2 3 = - t
Þ t = - 2 3 1 ( ) s
After attaining highest point
- = - 20 40 3 10t
i.e., - = - 2 2 3 t
or t = + 2 3 1 ( ) s
7. Average velocity =
Range
Time of flight
=
u g
u g
2
2
sin 2a
a /
/
sin
= ´
2
2
2
u
g
g
u
sin cos
sin
a a
a
= ucos a
8. Change in velocity
          = - - + ( sin ) ( sin ) u u a a
          = -2usina
          =2usina (downward)
9. Formulae for R T , and H
max
 will be same
if the projection point and the point where
the particle lands are same and lie on a
horizontal line.
10. r i j
®
= + - [ ( ) ]
^ ^
3 4 5
2
t t t m
y-coordinate will be zero when
4 5 0
2
t t - =
i.e., t = 0 s, 
4
5
 s
t = 0 belongs to the initial point of
projection of the particle.
\ r i
®
= 0
^
 m
i.e., x = 0 m
At t = 0.8 s,
r i
®
= 2.4
^
 m
i.e., x = 2.4 m
11. v u
x x
= = 10 m/s
      =10 m/s
        v u
x x
= = 10 m/s
        v v
x
= f cos
Þ    cosf=
v
v
x
          =
10
v
          =
10
10
[as, v =
20
2
 (given)]
          =1
i.e.,       f = ° 0
\ Speed will be half of its initial value at
the highest point where f = ° 0 .
Thus,       t
u
g
=
sin q
 =
°
=
20 60
10
3
sin
 s
52 | Mechanics-1
a
a
u – u sin a
u sin a
t = T
u cos a
+ u sin a u
60°
u = 20 m/s
v
x
v
f
u = 20 cos 60°
x
Introductory Exercise 4.2
1. Time of flight
T
u
g
=
- 2 sin ( )
cos
a b
b
                =
´ ° - °
°
( sin( )
cos
2 10 60 30
10 30
                          =
2
3
 s
Using,             v u at = +
  v u u
x x
= = ° cos 60
    = ´ = 10
1
2
5 m/s
v u g T
y y
= + - ( )
       = ° - u gT sin 60
        = - 10
3
2
10
2
3
 =
-
æ
è
ç
ö
ø
÷
10
3 4
2 3
= -
5
3
 m/s
                   v v v
x y net
= +
2 2
      = +
-
æ
è
ç
ö
ø
÷
5
5
3
2
2
= + 5 1
1
3
  
=
10
3
 m/s  
2. Component of velocity perpendicular to
plane
           =v
net
cosb
           = ´ °
10
3
30 cos
           = ´
10
3
3
2
           =5 m/s
3. Let the particle collide at time t.
x u t
1
= ( cos ) q
and x v t
2
=
\ d x x = -
2 1
= + ( cos ) v u t q
= + ° [ cos ] 10 10 2 45 t = 20 t
Using equation, s ut at = +
1
2
2
For vertical motion of particle 1 :
h u t g t - = + - 10
1
2
2
( sin ) ( ) q
i.e., h u t gt = + - 10
1
2
2
( sin ) q …(i)
or h t gt - + - 10 10
1
2
2
For the vertical motion of particle 2 :
20
1
2
2
- = h gt
i.e., h gt = - 20
1
2
2
…(ii)
Comparing Eqs. (i) and (ii),
10 10
1
2
20
1
2
2 2
+ - = - t g t g t
Þ t = 1 s
\ d = 20 m
4.                          u = 10 m/s
v = 5 2 m/s
Projectile Motion 53
h
v
u
q
x
2
x
1
d
10 m
2
1
q f
B A
d
v
u
a
b
u
a = 60°
b = 30°
2
u = 10 m/s
2
g = 10 m/s
v
x
v
net
v
y
Page 4


Introductory Exercise 4.1
1. A particle projected at any angle with
horizontal will always move in a plane
and thus projectile motion is a
2-dimensional motion. The statement is
thus false.
2. At high speed the projectile may go to a
place where acceleration due to gravity
has some different value and as such the
motion may not be uniform accelerated.
The statement is thus true.
3. See article 4.1.
4. u = 40 2 m/s, q = ° 45
As horizontal acceleration would be zero.
v u u
x x
= = = cos q 40 m/s
 s u t u t
x x
= = = ( cos ) q 80 m
A : position of particle at time = 0.
B : position of particle at time = t.
As vertical acceleration would be - g
          v u gt
y y
= -
= = u gt sinq
             = - 40 20
             =20 m/s
s u t gt
y y
= × -
1
2
2
  
= - ´ ´ = 80
1
2
10 2 60
2
 m
\ v v v
x y
= +
2 2
= + 40 20
2 2
= 20 5 m/s
tan f = =
v
v
y
x
20
40
i.e., f =
æ
è
ç
ö
ø
÷
-
tan
1
1
2
s s s
x y
= +
2 2
= + 80 60
2 2
= 100 m
tan a =
s
s
y
x
=
60
80
i.e., a =
æ
è
ç
ö
ø
÷
-
tan
1
3
4
5. s u t g t
y y
= + -
1
2
2
( )
or s u t gt
y
= - ( sin ) q
1
2
2
or 15 20
1
2
2
= - t gt
or t t
2
4 3 0 = + =
i.e., t = 1  s and 3 s
6. See figure to the answer to question no. 4.
u = 40 m/s, q = ° 60
\ u
x
= ° = 40 60 20 cos m/s
Projectile Motion
4
s
y
g
f
v
v
y
v
x
u
x
s
x
A
a
u
u
y
B
Thus, v u
x x
= = 20 m/s
As f = ° 45
tan f = =
v
v
y
x
1
\ y v
y x
= = 20 m/s
Thus, v v v v
x y x
= + =
2 2
2
= 20 2 m/s
Before reaching highest point
v u g t
y y
= + - ( )
\ 20 40 60 10 = ° - sin t
or 2 2 3 = - t
Þ t = - 2 3 1 ( ) s
After attaining highest point
- = - 20 40 3 10t
i.e., - = - 2 2 3 t
or t = + 2 3 1 ( ) s
7. Average velocity =
Range
Time of flight
=
u g
u g
2
2
sin 2a
a /
/
sin
= ´
2
2
2
u
g
g
u
sin cos
sin
a a
a
= ucos a
8. Change in velocity
          = - - + ( sin ) ( sin ) u u a a
          = -2usina
          =2usina (downward)
9. Formulae for R T , and H
max
 will be same
if the projection point and the point where
the particle lands are same and lie on a
horizontal line.
10. r i j
®
= + - [ ( ) ]
^ ^
3 4 5
2
t t t m
y-coordinate will be zero when
4 5 0
2
t t - =
i.e., t = 0 s, 
4
5
 s
t = 0 belongs to the initial point of
projection of the particle.
\ r i
®
= 0
^
 m
i.e., x = 0 m
At t = 0.8 s,
r i
®
= 2.4
^
 m
i.e., x = 2.4 m
11. v u
x x
= = 10 m/s
      =10 m/s
        v u
x x
= = 10 m/s
        v v
x
= f cos
Þ    cosf=
v
v
x
          =
10
v
          =
10
10
[as, v =
20
2
 (given)]
          =1
i.e.,       f = ° 0
\ Speed will be half of its initial value at
the highest point where f = ° 0 .
Thus,       t
u
g
=
sin q
 =
°
=
20 60
10
3
sin
 s
52 | Mechanics-1
a
a
u – u sin a
u sin a
t = T
u cos a
+ u sin a u
60°
u = 20 m/s
v
x
v
f
u = 20 cos 60°
x
Introductory Exercise 4.2
1. Time of flight
T
u
g
=
- 2 sin ( )
cos
a b
b
                =
´ ° - °
°
( sin( )
cos
2 10 60 30
10 30
                          =
2
3
 s
Using,             v u at = +
  v u u
x x
= = ° cos 60
    = ´ = 10
1
2
5 m/s
v u g T
y y
= + - ( )
       = ° - u gT sin 60
        = - 10
3
2
10
2
3
 =
-
æ
è
ç
ö
ø
÷
10
3 4
2 3
= -
5
3
 m/s
                   v v v
x y net
= +
2 2
      = +
-
æ
è
ç
ö
ø
÷
5
5
3
2
2
= + 5 1
1
3
  
=
10
3
 m/s  
2. Component of velocity perpendicular to
plane
           =v
net
cosb
           = ´ °
10
3
30 cos
           = ´
10
3
3
2
           =5 m/s
3. Let the particle collide at time t.
x u t
1
= ( cos ) q
and x v t
2
=
\ d x x = -
2 1
= + ( cos ) v u t q
= + ° [ cos ] 10 10 2 45 t = 20 t
Using equation, s ut at = +
1
2
2
For vertical motion of particle 1 :
h u t g t - = + - 10
1
2
2
( sin ) ( ) q
i.e., h u t gt = + - 10
1
2
2
( sin ) q …(i)
or h t gt - + - 10 10
1
2
2
For the vertical motion of particle 2 :
20
1
2
2
- = h gt
i.e., h gt = - 20
1
2
2
…(ii)
Comparing Eqs. (i) and (ii),
10 10
1
2
20
1
2
2 2
+ - = - t g t g t
Þ t = 1 s
\ d = 20 m
4.                          u = 10 m/s
v = 5 2 m/s
Projectile Motion 53
h
v
u
q
x
2
x
1
d
10 m
2
1
q f
B A
d
v
u
a
b
u
a = 60°
b = 30°
2
u = 10 m/s
2
g = 10 m/s
v
x
v
net
v
y
q = ° 30
f = ° 45
d = 15 m
Let the particles meet (or are in the same
vertical time t).
\ d u t v t = + f ( cos ) ( cos ) q
Þ 15 10 30 5 2 45 = ° + ° ( cos cos ) t
or 15 5 3 5 = + ( ) t
or t =
+
3
3 1
 s
= 1.009 s
Now, let us find time of flight of A and B
T
u
g
A
=
2 sin q
= 1 s
As T t
A
< , particle A will touch ground
before the expected time t of collision.
\      Ans : NO.
5. For range to be maximum
  a
p b
= +
4 2
          = +
p p
4
6
2
/
=
p
3
   = ° 60
6. At point A velocity ( ) v
®
 of the particle will
be parallel to the inclined plane.
   u = 40 m/s
a = ° 60
b = ° 30
        g = 10 m/s
2
\ f = b
v u u
x x
= = cos a
v v v
x
= f = cos cos b
or u v cos cos a b =
Þ v
u
= =
°
°
cos
cos
cos
cos
a
b
40 60
30
=
æ
è
ç
ö
ø
÷ 40
1
2
3
2
 
=
40
3
 m/s
7. (a) At time t, vertical displacement of A
= Vertical displacement of B
      ( sin ) v t gt v t gt
A B
q - = -
1
2
1
2
2 2
i.e.,                v v
A B
sin q =
sin q =
v
v
B
A
            = =
10
20
1
2
\     q = ° 30
(b) x v t
A
= ( cos ) q
        = ° ´ ( cos ) 20 30
1
2
         = 5 3 m
54 | Mechanics-1
a
b = 30° =
u
p
6
a
b
u
A
f A
v
time t
v = 10 m/s
B
2
a = – 10 m/s
v = 20 m/s
A
A
B
x
Page 5


Introductory Exercise 4.1
1. A particle projected at any angle with
horizontal will always move in a plane
and thus projectile motion is a
2-dimensional motion. The statement is
thus false.
2. At high speed the projectile may go to a
place where acceleration due to gravity
has some different value and as such the
motion may not be uniform accelerated.
The statement is thus true.
3. See article 4.1.
4. u = 40 2 m/s, q = ° 45
As horizontal acceleration would be zero.
v u u
x x
= = = cos q 40 m/s
 s u t u t
x x
= = = ( cos ) q 80 m
A : position of particle at time = 0.
B : position of particle at time = t.
As vertical acceleration would be - g
          v u gt
y y
= -
= = u gt sinq
             = - 40 20
             =20 m/s
s u t gt
y y
= × -
1
2
2
  
= - ´ ´ = 80
1
2
10 2 60
2
 m
\ v v v
x y
= +
2 2
= + 40 20
2 2
= 20 5 m/s
tan f = =
v
v
y
x
20
40
i.e., f =
æ
è
ç
ö
ø
÷
-
tan
1
1
2
s s s
x y
= +
2 2
= + 80 60
2 2
= 100 m
tan a =
s
s
y
x
=
60
80
i.e., a =
æ
è
ç
ö
ø
÷
-
tan
1
3
4
5. s u t g t
y y
= + -
1
2
2
( )
or s u t gt
y
= - ( sin ) q
1
2
2
or 15 20
1
2
2
= - t gt
or t t
2
4 3 0 = + =
i.e., t = 1  s and 3 s
6. See figure to the answer to question no. 4.
u = 40 m/s, q = ° 60
\ u
x
= ° = 40 60 20 cos m/s
Projectile Motion
4
s
y
g
f
v
v
y
v
x
u
x
s
x
A
a
u
u
y
B
Thus, v u
x x
= = 20 m/s
As f = ° 45
tan f = =
v
v
y
x
1
\ y v
y x
= = 20 m/s
Thus, v v v v
x y x
= + =
2 2
2
= 20 2 m/s
Before reaching highest point
v u g t
y y
= + - ( )
\ 20 40 60 10 = ° - sin t
or 2 2 3 = - t
Þ t = - 2 3 1 ( ) s
After attaining highest point
- = - 20 40 3 10t
i.e., - = - 2 2 3 t
or t = + 2 3 1 ( ) s
7. Average velocity =
Range
Time of flight
=
u g
u g
2
2
sin 2a
a /
/
sin
= ´
2
2
2
u
g
g
u
sin cos
sin
a a
a
= ucos a
8. Change in velocity
          = - - + ( sin ) ( sin ) u u a a
          = -2usina
          =2usina (downward)
9. Formulae for R T , and H
max
 will be same
if the projection point and the point where
the particle lands are same and lie on a
horizontal line.
10. r i j
®
= + - [ ( ) ]
^ ^
3 4 5
2
t t t m
y-coordinate will be zero when
4 5 0
2
t t - =
i.e., t = 0 s, 
4
5
 s
t = 0 belongs to the initial point of
projection of the particle.
\ r i
®
= 0
^
 m
i.e., x = 0 m
At t = 0.8 s,
r i
®
= 2.4
^
 m
i.e., x = 2.4 m
11. v u
x x
= = 10 m/s
      =10 m/s
        v u
x x
= = 10 m/s
        v v
x
= f cos
Þ    cosf=
v
v
x
          =
10
v
          =
10
10
[as, v =
20
2
 (given)]
          =1
i.e.,       f = ° 0
\ Speed will be half of its initial value at
the highest point where f = ° 0 .
Thus,       t
u
g
=
sin q
 =
°
=
20 60
10
3
sin
 s
52 | Mechanics-1
a
a
u – u sin a
u sin a
t = T
u cos a
+ u sin a u
60°
u = 20 m/s
v
x
v
f
u = 20 cos 60°
x
Introductory Exercise 4.2
1. Time of flight
T
u
g
=
- 2 sin ( )
cos
a b
b
                =
´ ° - °
°
( sin( )
cos
2 10 60 30
10 30
                          =
2
3
 s
Using,             v u at = +
  v u u
x x
= = ° cos 60
    = ´ = 10
1
2
5 m/s
v u g T
y y
= + - ( )
       = ° - u gT sin 60
        = - 10
3
2
10
2
3
 =
-
æ
è
ç
ö
ø
÷
10
3 4
2 3
= -
5
3
 m/s
                   v v v
x y net
= +
2 2
      = +
-
æ
è
ç
ö
ø
÷
5
5
3
2
2
= + 5 1
1
3
  
=
10
3
 m/s  
2. Component of velocity perpendicular to
plane
           =v
net
cosb
           = ´ °
10
3
30 cos
           = ´
10
3
3
2
           =5 m/s
3. Let the particle collide at time t.
x u t
1
= ( cos ) q
and x v t
2
=
\ d x x = -
2 1
= + ( cos ) v u t q
= + ° [ cos ] 10 10 2 45 t = 20 t
Using equation, s ut at = +
1
2
2
For vertical motion of particle 1 :
h u t g t - = + - 10
1
2
2
( sin ) ( ) q
i.e., h u t gt = + - 10
1
2
2
( sin ) q …(i)
or h t gt - + - 10 10
1
2
2
For the vertical motion of particle 2 :
20
1
2
2
- = h gt
i.e., h gt = - 20
1
2
2
…(ii)
Comparing Eqs. (i) and (ii),
10 10
1
2
20
1
2
2 2
+ - = - t g t g t
Þ t = 1 s
\ d = 20 m
4.                          u = 10 m/s
v = 5 2 m/s
Projectile Motion 53
h
v
u
q
x
2
x
1
d
10 m
2
1
q f
B A
d
v
u
a
b
u
a = 60°
b = 30°
2
u = 10 m/s
2
g = 10 m/s
v
x
v
net
v
y
q = ° 30
f = ° 45
d = 15 m
Let the particles meet (or are in the same
vertical time t).
\ d u t v t = + f ( cos ) ( cos ) q
Þ 15 10 30 5 2 45 = ° + ° ( cos cos ) t
or 15 5 3 5 = + ( ) t
or t =
+
3
3 1
 s
= 1.009 s
Now, let us find time of flight of A and B
T
u
g
A
=
2 sin q
= 1 s
As T t
A
< , particle A will touch ground
before the expected time t of collision.
\      Ans : NO.
5. For range to be maximum
  a
p b
= +
4 2
          = +
p p
4
6
2
/
=
p
3
   = ° 60
6. At point A velocity ( ) v
®
 of the particle will
be parallel to the inclined plane.
   u = 40 m/s
a = ° 60
b = ° 30
        g = 10 m/s
2
\ f = b
v u u
x x
= = cos a
v v v
x
= f = cos cos b
or u v cos cos a b =
Þ v
u
= =
°
°
cos
cos
cos
cos
a
b
40 60
30
=
æ
è
ç
ö
ø
÷ 40
1
2
3
2
 
=
40
3
 m/s
7. (a) At time t, vertical displacement of A
= Vertical displacement of B
      ( sin ) v t gt v t gt
A B
q - = -
1
2
1
2
2 2
i.e.,                v v
A B
sin q =
sin q =
v
v
B
A
            = =
10
20
1
2
\     q = ° 30
(b) x v t
A
= ( cos ) q
        = ° ´ ( cos ) 20 30
1
2
         = 5 3 m
54 | Mechanics-1
a
b = 30° =
u
p
6
a
b
u
A
f A
v
time t
v = 10 m/s
B
2
a = – 10 m/s
v = 20 m/s
A
A
B
x
AIEEE Cor ner
Subjective Questions (Level 1)
1. (a) R
u
g
= =
°
2 2
2 20 2 90
10
sin ( ) sin q
= 80 m
   H
u
g
= =
°
2 2 2 2
2
20 2 45
10
sin ( ) sin q
= 40 m
   T
u
g
= =
° 2 2 20 2 45
10
sin ( ) sin q
= 4 s
(b) u i j
®
= + ( )
^ ^
20 20 m/s and a j
®
= - 10
^
 m/s
2
- = +
® ® ®
v u a t
= + + - ( ) ( )
^ ^ ^
20 20 10 1 i j j (at t = 1 s)
= - ( )
^ ^
20 10 i j m/s
(c) Time of flight 
T
u
g
=
2 sin q
=
´ ´
æ
è
ç
ö
ø
÷
2 20 2
1
2
10
= 4 s
(c) \ Velocity of particle at the time of
collision with ground.
= + + - ( ) ( )
^ ^ ^
20 20 10 4 i j j
= - ( )
^ ^
20 20 i j m/s
2. (a) s ut at = +
1
2
2
\ ( ) ( ) ( ) - = + + - 40 20
1
2
10
2
T T
or 5 20 40 0
2
T T - - =
or   T T
2
4 8 0 - - =
\ T =
- - + - - - ( ) ( ) ( )( )
( )
4 4 4 1 8
2 1
2
Leaving - ive sign which is not positive.
                =
+ 4 48
2
 =
+ 4 4 3
2
  = + ( ) 2 2 3 s
              =5.46 s
       R = ´ = 20 5.46 109.2 m
(b) s ut at = +
1
2
2
( ) ( ) - = ´ + - 40
1
2
10
2
0 T T
Þ T = 2 2 s
= 2.83 s
R = ´ 2.83 20
= 56.6 m
(c) s ut at = +
1
2
2
( ) ( ) ( ) - = - + - 40 20
1
2
10
2
T T
or 5 20 40 0
2
T T + - =
or T T
2
4 8 0 + - =
\ T =
- + - - 4 4 4 1 8
2 1
2
( ) ( ) ( )
( )
              =
- + 4 4 3
2
 = - + 2 2 3 = 1.46 s
Projectile Motion 55
R
– 40 m
+
T
2
a = – 10 m/s
45°
20Ö2 m/s
20 m/s
R
– 40 m
+
T
20 m/s
2
a = – 10 m/s
R
T
– 40 m
20 Ö2 m/s
Read More
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