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CHAPTER NO.7 
CIRCULAR MOTION 
7.1 Projectile Motion (an example of circular motion) 
Terms: 
7.1.1 Projectile (Nothing but particle): 
It is a particle or object projected in the space at an inclination to the direction of gravity, moving 
under the combined effect of vertical and horizontal forces. 
1) Trajectory: 
It is a path, traced by a projectile in the space. 
2) Point of Projection: 
It is the initial point of the particle (see Fig. 1 Point A) 
3) Angle of inclination or Projection: 
The angle, with a horizontal, at which a projectile is projected (see Fig. 1) 
4) Horizontal Range or Range (R) : 
The distance between the point of projection and the point where projectile strikes the 
ground (see Fig. 1) 
5) Time of Flight: 
The total time taken by the projectile, to reach the maximum height i.e. hmax and to return 
back to the ground. 
6) Velocity of projection: 
The velocity with which, a projectile is projected (see Fig. 1) 
Chapter No. 7 Circular Motion Page 1
Fig. 1 
7.2 Study of a projectile: 
1) Generally, the velocity of the projection ‘u’ split up in to or resolved into two 
components ‘u cos α’ and ‘u sin α’. 
i. e. Vertical component V = u sin α 
and Horizontal component H = u cos α 
2) The component in vertical direction is subjected to retardation due to gravity. Therefore, 
this vertical velocity ‘u sin α’ is subjected to retardation for first half of trajectory before 
attaining maximum height i.e. hmax and for second half it is subjected to acceleration due 
to gravity, before coming to rest. 
3) The component in horizontal direction, there is no retardation or acceleration due to 
gravity. Therefore, this horizontal velocity ‘u cos α’ is remain constant throughout the 
motion. 
For first half of trajectory ---------- { 
푣 = 푢 sin 훼 − 푔푡 
푣 2 = 푢2 sin2 훼 − 2푔푠 
푠 = ( 푢 sin 훼 ) 푡 − (1 
2 
푔 푡2) 
Chapter No. 7 Circular Motion Page 2
For second half of trajectory ---------- { 
푣 = 푢 sin 훼 + 푔푡 
푣2 = 푢2 sin2 훼 + 2푔푠 
푠 = ( 푢 sin 훼 ) 푡 + (1 
2 
푔 푡2) 
4) Thus, the time taken by the body or particle or projectile to reach the ground, is 
calculated by vertical component (u sin α ) of the velocity and the horizontal component 
of the velocity ( u cos α). 
5) Mathematically, 
퐻표푟푖푧표푛푡푎푙 푅푎푛푔푒 (푅) 
[ 
표푟 
퐻표푟푖푧표푛푡푎푙 퐷푖푠푡푎푛푐푒 
퐶표푛푠푡푎푛푡 퐻표푟푖푧표푛푡푎푙 푣푒푙표푐푖푡푦 
] = [ 
(푢 cos 훼 ) 
푇푖푚푒 
(푡) 
] 푥 [ 
] 
Fig. 2 
7.3 Derivation of the expression or equation for the path of a projectile: 
Consider a particle projected upward from point A, at an angle α, with the horizontal, with an 
initial velocity u m/s as shown in Fig. 1 below. 
Now, the velocity of projection u can be split up into two components i.e. u cos α and u sin α. 
Chapter No. 7 Circular Motion Page 3
Fig. 3 
Consider any point as the position of the particle, after time t seconds with co-ordinates x and y 
as shown in Fig. 1 
푦 = ( 푢 sin 훼 ) 푡 − (1 
2 
푔 푡2) ------------------- (1) 
And 푥 = ( 푢 cos 훼 ) 푡 ----------- (2) 
Therefore, 푡 = 푥 
푢 cos 훼 
Substituting in (1) 
푦 = ( 푢 sin 훼 ) 
푥 
푢 cos 훼 
− ( 
1 
2 
푔 
푥2 
) 
푢2 cos2 훼 
푦 = 푥 tan 훼 − 푔 푥2 
2 푐표푠2 푢2 ---------- (3) 
Equation (3) is known as equation of trajectory 
7.4 Derivation of expression for the time of flight of the projectile on a horizontal plane: 
We have, 푠 = 푢푡 + 1 
2 
푎 푡2 
Chapter No. 7 Circular Motion Page 4
푦 = ( 푢 sin 훼 ) 푡 − (1 
2 
푔 푡2) ------------------- (1) 
i.e. s = y, u = u sin α and a = g 
When the projectile p is at point B, y = 0 
Substituting, the value of y in eqn (1) 
0 = ( 푢 sin 훼 ) 푡 − ( 
1 
2 
푔 푡2) 
( 푢 sin 훼 ) 푡 = ( 
1 
2 
푔 푡2) 
( 푢 sin 훼 ) = ( 
1 
2 
푔 푡) 
푡 = 
2 푢 sin 훼 
푔 
i.e. 1) time required to reach the maximum height 
푡 
2 
= 
푢 sin 훼 
푔 
Chapter No. 7 Circular Motion Page 5
7.5 Derivation of the expression for horizontal range (R) of the projectile: 
We know, 
Horizontal component of velocity = u cos α 
And 
Time of flight, 
푡 = 
2 푢 sin 훼 
푔 
Therefore, 
[Horizontal range ] = [퐻표푟푖푧표푛푡푎푙 푣푒푙표푐푖푡푦] 푥 [Time of flight] 
= [u cos α] 푥 [2 푢 sin 훼 
푔 
] 
푅 = 
sin 2 훼 푢2 
푔 
Note: 
The range will be maximum when sin 2α = 1 
Chapter No. 7 Circular Motion Page 6
2α = 900 
α = 450 
Rmax = 
푢2 
푔 
7.6 Derivation of expression for the maximum height of a projectile on a horizontal plane: 
We have, the vertical component of initial velocity u is 
v = u sin α 
and vertical component of final velocity 
v = 0 
Therefore, 
Average velocity = 
푢 sin 훼 +0 
2 
= 
푢 sin 훼 
2 
Now, 
[푣푒푟푡푖푐푎푙 푑푖푠푡푎푛푐푒] = [퐴푣푒푟푎푔푒 푣푒푟푡푖푐푎푙 푣푒푙표푐푖푡푦] x [푡푖푚푒] 
Chapter No. 7 Circular Motion Page 7
= 
푢 sin 훼 
2 
푥 푢 sin 훼 
푔 
= 
푠푖푛2 훼 푢2 
2 푔 
Where, 
푢 sin 훼 
푔 
= is the time required by the projectile to reach the maximum height 
H = 푠푖푛2 훼 푢2 
2 푔 
Questions:- 
1. A projectile is fired upwards at an angle of 300 with a velocity of 40 m/s. Calculate the 
time taken by the projectile to reach the ground after firing. 
2. If a particle is projected inside a horizontal tunnel which is 5 m high with a velocity of 
60 m/s, find the angle of projection and the greatest possible range. 
3. The range of projectile on a horizontal plane is 240 m and the time of flight is 8 sec. Find 
the initial velocity and the angle of projection. 
4. A body is projected at such an angle that the horizontal range is three times the greatest 
height. Find the angle of projection. 
5. Find the least initial velocity which a projectile may have, so that it may clear a wall 3.6 
m high and 4.8 m distant (from the point of projection) and strike the horizontal plane 
through the foot of the wall at a distance 3.6 m beyond the wall. The point of projection is 
at the same level as the wall. 
6. Find the angle of projection at which the horizontal range and the maximum height of a 
projectile are equal. May 2010 (06 MKS) 
7. A particle is projected in air with a velocity 100 m/s and at an angle of 300 with the 
horizontal. Find: 
1) The horizontal range 
2) The maximum height by the particle 
3) And the time of flight 
Ans: R = 882.8 m, Hmax = 127.42 m, t = 10.19 sec 
Chapter No. 7 Circular Motion Page 8
8. A projectile is fired with an initial velocity of 250 m/s at a target located at a horizontal 
distance of 4 km and the vertical distance of 700 m above the gun. Determine the value of 
firing angle to hit the target. (Neglect air resistance). 
Ans: α = 68.820 and 31.080 
9. A projectile is aimed to a mark on a horizontal plane through the point of projection and 
fall 12 m short of the mark and with an angle of projection 150. While it overshoots the 
mark by 24 m when the angle of projection to hit the mark. Assume no air resistance. 
Take the velocity of projection is constant in all cases. 
Ans: R = 48 m, u = 26.57 m, α = 20.550 
Case I) Projectile projected horizontally from certain height 
a) Considering vertical motion of projectile 
We have, 
S = 퐮퐭 + 
ퟏ 
ퟐ 
퐚퐭ퟐ 
Here, 
S = y = vertically downward displacement 
u = initial velocity = u sin α = 0 
a = acceleration due to gravity = g = 9.81 m/s2 
∴ y = 
ퟏ 
ퟐ 
퐠 퐭ퟐ---------- (1) 
b) Considering horizontal motion of projectile 
[Horizontal range ] = [퐻표푟푖푧표푛푡푎푙 푣푒푙표푐푖푡푦] 푥 [Time of flight] 
x = u x t --------- (2) 
Chapter No. 7 Circular Motion Page 9
Problems: 
1) An aircraft moving horizontally at a speed of 1000 kmph at a height of 1500 m releases a 
bomb which hits the target. 
Find: 
1) Time required for the bomb to reach the target on ground 
2) The horizontal distance of the aircraft from the target when it releases the 
bomb. (May 2007 8 MKS) 
Ans: t = 17.49 sec and x = 4857.66 m 
2) Solve the same problem when y = 2000 m and u = 150 m/s. 
Ans: t = 20.2 sec and x = 3030m 
3) Find out u = ? when y = 100 m and x = 200 m 
Ans: t = 4.515 sec and u = 44.3 m/s 
4) Find y = ? and u = ? when time required to strike the ground is 15 sec when x = 50 m. 
Ans: y = 1103.625 m and u = 3.33 m/s 
5) A person wants to jump over a ditch as shown in Fig. below. Find the max. velocity with 
which he should jump? (May 2006 6 MKS) 
Ans: t = 0.638 sec and u = 4.702 m/s 
Chapter No. 7 Circular Motion Page 10
Case II) Projectile projected at an angle α from certain height (say y) 
a) Considering vertical motion of projectile 
We have, 
S = 퐮퐭 + 
ퟏ 
ퟐ 
퐚퐭ퟐ 
Here, 
S = y = vertically downward displacement 
u = initial vertical velocity = u sin α = 0 
a = acceleration due to gravity = g = 9.81 m/s2 
∴ − y = (u sin α) – 
ퟏ 
ퟐ 
퐠 퐭ퟐ 
y = - (u sin α) + 
ퟏ 
ퟐ 
퐠 퐭ퟐ---------- (1) 
b) Considering horizontal motion of projectile 
[Horizontal range ] = [퐻표푟푖푧표푛푡푎푙 푣푒푙표푐푖푡푦] 푥 [Time of flight] 
x = u cos α x t --------- (2) 
Velocity with which it strikes the ground (VB): 
VB = u = initial velocity 
Chapter No. 7 Circular Motion Page 11
VH = u cos α 
VV = u sin α 
We have, 
퐯 = 퐮 + 퐚퐭 
-VV = u sin α – gt 
VV = -u sin α + gt 
ퟐ + 푽푽 
VB = √ 푽푯 
ퟐ--------------- (velocity of hit) 
Direction of hit: 
θ = tan-1 ( 
푽푽 
푽푯 
) ------------ (with the horizontal) 
Problems: 
1. A body is projected from the top of the tower 40 m high and strikes the ground after 10 
seconds at a point 400 m from the base of the tower. Determine the velocity and angle of 
projection. Also determine the maximum height attained by the body above ground level. 
Dec. 2009 (10 MKS) 
Ans: u = 60.24 m/s and Hmax = 143.43 m 
2. A cricket ball is thrown by a fielder from a height of 2 , at an angle of elevation 300 to the 
horizontal with an initial velocity of 20 m/s, hits the wickets at a height of 0.5 m from the 
ground. How far the fielder from the wicket? (May 2004 12 MKS) 
Ans: t = 2.181 sec, x = 37.77 m 
3. A solider fires a bullet at an angle of elevation 300 from his position on the top of hill to 
strike a target which is 60 m lower than the position of the solider. The initial velocity of 
the bullet is 75 m/s. Calculate: 
a) Hmax to which bullet rise above horizontal 
b) The actual velocity with which it will strike the target 
c) Total time required by for the flight of the bullet 
Chapter No. 7 Circular Motion Page 12
Ans: Hmax = 131.74 m, t = 9.01 sec, vB = 82.456 m/s, θ = 38.020 
4. A projectile fired from the edge of a 150 m high cliff with an initial velocity of 180 m/s at 
an angle of elevation of 300 with the horizontal. Neglecting air resistance. Find: 
1) The greatest elevation above the ground reached by the projectile and 
2) Horizontal distance from gun to the point, where the projectile strikes the ground. 
Ans: Hmax = 563.3 m, t = 19.9 sec, R = 3.012 km 
Case III) Projectile projected at an angle α from certain height to the downward (say y) 
c) Considering vertical motion of projectile 
We have, 
S = 퐮퐭 + 
ퟏ 
ퟐ 
퐚퐭ퟐ 
Here, 
S = - y = vertically downward displacement 
u = initial vertical velocity = - u sin α 
a = acceleration due to gravity = - g = 9.81 m/s2 
∴ − y = ( - u sin α) – 
ퟏ 
ퟐ 
퐠 퐭ퟐ 
y = (u sin α) + 
ퟏ 
ퟐ 
퐠 퐭ퟐ---------- (1) 
d) Considering horizontal motion of projectile 
[Horizontal range ] = [퐻표푟푖푧표푛푡푎푙 푣푒푙표푐푖푡푦] 푥 [Time of flight] 
Chapter No. 7 Circular Motion Page 13
x = u cos α x t --------- (2) 
Velocity with which it strikes the ground (VB): 
VB = u = initial velocity 
VH = u cos α 
VV = u sin α 
We have, 
퐯 = 퐮 + 퐚퐭 
-VV = - u sin α – gt 
VV = u sin α + gt 
ퟐ + 푽푽 
VB = √ 푽푯 
ퟐ--------------- (velocity of hit) 
Direction of hit: 
θ = tan-1 ( 
푽푽 
푽푯 
) ------------ (with the horizontal) 
Problems: 
1) A stone is thrown from the top of the tower 30 m high at an angle of 400 with horizontal 
with velocity 40 kmph. Find when and where the stone will hit a horizontal ground. Also 
find the velocity and direction of striking. 
Ans: t = 1.85 sec, x = 15.74 m, Vb = 26.67 m/s, θ = 71.390 
Chapter No. 7 Circular Motion Page 14

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Projectile Motion

  • 1. CHAPTER NO.7 CIRCULAR MOTION 7.1 Projectile Motion (an example of circular motion) Terms: 7.1.1 Projectile (Nothing but particle): It is a particle or object projected in the space at an inclination to the direction of gravity, moving under the combined effect of vertical and horizontal forces. 1) Trajectory: It is a path, traced by a projectile in the space. 2) Point of Projection: It is the initial point of the particle (see Fig. 1 Point A) 3) Angle of inclination or Projection: The angle, with a horizontal, at which a projectile is projected (see Fig. 1) 4) Horizontal Range or Range (R) : The distance between the point of projection and the point where projectile strikes the ground (see Fig. 1) 5) Time of Flight: The total time taken by the projectile, to reach the maximum height i.e. hmax and to return back to the ground. 6) Velocity of projection: The velocity with which, a projectile is projected (see Fig. 1) Chapter No. 7 Circular Motion Page 1
  • 2. Fig. 1 7.2 Study of a projectile: 1) Generally, the velocity of the projection ‘u’ split up in to or resolved into two components ‘u cos α’ and ‘u sin α’. i. e. Vertical component V = u sin α and Horizontal component H = u cos α 2) The component in vertical direction is subjected to retardation due to gravity. Therefore, this vertical velocity ‘u sin α’ is subjected to retardation for first half of trajectory before attaining maximum height i.e. hmax and for second half it is subjected to acceleration due to gravity, before coming to rest. 3) The component in horizontal direction, there is no retardation or acceleration due to gravity. Therefore, this horizontal velocity ‘u cos α’ is remain constant throughout the motion. For first half of trajectory ---------- { 푣 = 푢 sin 훼 − 푔푡 푣 2 = 푢2 sin2 훼 − 2푔푠 푠 = ( 푢 sin 훼 ) 푡 − (1 2 푔 푡2) Chapter No. 7 Circular Motion Page 2
  • 3. For second half of trajectory ---------- { 푣 = 푢 sin 훼 + 푔푡 푣2 = 푢2 sin2 훼 + 2푔푠 푠 = ( 푢 sin 훼 ) 푡 + (1 2 푔 푡2) 4) Thus, the time taken by the body or particle or projectile to reach the ground, is calculated by vertical component (u sin α ) of the velocity and the horizontal component of the velocity ( u cos α). 5) Mathematically, 퐻표푟푖푧표푛푡푎푙 푅푎푛푔푒 (푅) [ 표푟 퐻표푟푖푧표푛푡푎푙 퐷푖푠푡푎푛푐푒 퐶표푛푠푡푎푛푡 퐻표푟푖푧표푛푡푎푙 푣푒푙표푐푖푡푦 ] = [ (푢 cos 훼 ) 푇푖푚푒 (푡) ] 푥 [ ] Fig. 2 7.3 Derivation of the expression or equation for the path of a projectile: Consider a particle projected upward from point A, at an angle α, with the horizontal, with an initial velocity u m/s as shown in Fig. 1 below. Now, the velocity of projection u can be split up into two components i.e. u cos α and u sin α. Chapter No. 7 Circular Motion Page 3
  • 4. Fig. 3 Consider any point as the position of the particle, after time t seconds with co-ordinates x and y as shown in Fig. 1 푦 = ( 푢 sin 훼 ) 푡 − (1 2 푔 푡2) ------------------- (1) And 푥 = ( 푢 cos 훼 ) 푡 ----------- (2) Therefore, 푡 = 푥 푢 cos 훼 Substituting in (1) 푦 = ( 푢 sin 훼 ) 푥 푢 cos 훼 − ( 1 2 푔 푥2 ) 푢2 cos2 훼 푦 = 푥 tan 훼 − 푔 푥2 2 푐표푠2 푢2 ---------- (3) Equation (3) is known as equation of trajectory 7.4 Derivation of expression for the time of flight of the projectile on a horizontal plane: We have, 푠 = 푢푡 + 1 2 푎 푡2 Chapter No. 7 Circular Motion Page 4
  • 5. 푦 = ( 푢 sin 훼 ) 푡 − (1 2 푔 푡2) ------------------- (1) i.e. s = y, u = u sin α and a = g When the projectile p is at point B, y = 0 Substituting, the value of y in eqn (1) 0 = ( 푢 sin 훼 ) 푡 − ( 1 2 푔 푡2) ( 푢 sin 훼 ) 푡 = ( 1 2 푔 푡2) ( 푢 sin 훼 ) = ( 1 2 푔 푡) 푡 = 2 푢 sin 훼 푔 i.e. 1) time required to reach the maximum height 푡 2 = 푢 sin 훼 푔 Chapter No. 7 Circular Motion Page 5
  • 6. 7.5 Derivation of the expression for horizontal range (R) of the projectile: We know, Horizontal component of velocity = u cos α And Time of flight, 푡 = 2 푢 sin 훼 푔 Therefore, [Horizontal range ] = [퐻표푟푖푧표푛푡푎푙 푣푒푙표푐푖푡푦] 푥 [Time of flight] = [u cos α] 푥 [2 푢 sin 훼 푔 ] 푅 = sin 2 훼 푢2 푔 Note: The range will be maximum when sin 2α = 1 Chapter No. 7 Circular Motion Page 6
  • 7. 2α = 900 α = 450 Rmax = 푢2 푔 7.6 Derivation of expression for the maximum height of a projectile on a horizontal plane: We have, the vertical component of initial velocity u is v = u sin α and vertical component of final velocity v = 0 Therefore, Average velocity = 푢 sin 훼 +0 2 = 푢 sin 훼 2 Now, [푣푒푟푡푖푐푎푙 푑푖푠푡푎푛푐푒] = [퐴푣푒푟푎푔푒 푣푒푟푡푖푐푎푙 푣푒푙표푐푖푡푦] x [푡푖푚푒] Chapter No. 7 Circular Motion Page 7
  • 8. = 푢 sin 훼 2 푥 푢 sin 훼 푔 = 푠푖푛2 훼 푢2 2 푔 Where, 푢 sin 훼 푔 = is the time required by the projectile to reach the maximum height H = 푠푖푛2 훼 푢2 2 푔 Questions:- 1. A projectile is fired upwards at an angle of 300 with a velocity of 40 m/s. Calculate the time taken by the projectile to reach the ground after firing. 2. If a particle is projected inside a horizontal tunnel which is 5 m high with a velocity of 60 m/s, find the angle of projection and the greatest possible range. 3. The range of projectile on a horizontal plane is 240 m and the time of flight is 8 sec. Find the initial velocity and the angle of projection. 4. A body is projected at such an angle that the horizontal range is three times the greatest height. Find the angle of projection. 5. Find the least initial velocity which a projectile may have, so that it may clear a wall 3.6 m high and 4.8 m distant (from the point of projection) and strike the horizontal plane through the foot of the wall at a distance 3.6 m beyond the wall. The point of projection is at the same level as the wall. 6. Find the angle of projection at which the horizontal range and the maximum height of a projectile are equal. May 2010 (06 MKS) 7. A particle is projected in air with a velocity 100 m/s and at an angle of 300 with the horizontal. Find: 1) The horizontal range 2) The maximum height by the particle 3) And the time of flight Ans: R = 882.8 m, Hmax = 127.42 m, t = 10.19 sec Chapter No. 7 Circular Motion Page 8
  • 9. 8. A projectile is fired with an initial velocity of 250 m/s at a target located at a horizontal distance of 4 km and the vertical distance of 700 m above the gun. Determine the value of firing angle to hit the target. (Neglect air resistance). Ans: α = 68.820 and 31.080 9. A projectile is aimed to a mark on a horizontal plane through the point of projection and fall 12 m short of the mark and with an angle of projection 150. While it overshoots the mark by 24 m when the angle of projection to hit the mark. Assume no air resistance. Take the velocity of projection is constant in all cases. Ans: R = 48 m, u = 26.57 m, α = 20.550 Case I) Projectile projected horizontally from certain height a) Considering vertical motion of projectile We have, S = 퐮퐭 + ퟏ ퟐ 퐚퐭ퟐ Here, S = y = vertically downward displacement u = initial velocity = u sin α = 0 a = acceleration due to gravity = g = 9.81 m/s2 ∴ y = ퟏ ퟐ 퐠 퐭ퟐ---------- (1) b) Considering horizontal motion of projectile [Horizontal range ] = [퐻표푟푖푧표푛푡푎푙 푣푒푙표푐푖푡푦] 푥 [Time of flight] x = u x t --------- (2) Chapter No. 7 Circular Motion Page 9
  • 10. Problems: 1) An aircraft moving horizontally at a speed of 1000 kmph at a height of 1500 m releases a bomb which hits the target. Find: 1) Time required for the bomb to reach the target on ground 2) The horizontal distance of the aircraft from the target when it releases the bomb. (May 2007 8 MKS) Ans: t = 17.49 sec and x = 4857.66 m 2) Solve the same problem when y = 2000 m and u = 150 m/s. Ans: t = 20.2 sec and x = 3030m 3) Find out u = ? when y = 100 m and x = 200 m Ans: t = 4.515 sec and u = 44.3 m/s 4) Find y = ? and u = ? when time required to strike the ground is 15 sec when x = 50 m. Ans: y = 1103.625 m and u = 3.33 m/s 5) A person wants to jump over a ditch as shown in Fig. below. Find the max. velocity with which he should jump? (May 2006 6 MKS) Ans: t = 0.638 sec and u = 4.702 m/s Chapter No. 7 Circular Motion Page 10
  • 11. Case II) Projectile projected at an angle α from certain height (say y) a) Considering vertical motion of projectile We have, S = 퐮퐭 + ퟏ ퟐ 퐚퐭ퟐ Here, S = y = vertically downward displacement u = initial vertical velocity = u sin α = 0 a = acceleration due to gravity = g = 9.81 m/s2 ∴ − y = (u sin α) – ퟏ ퟐ 퐠 퐭ퟐ y = - (u sin α) + ퟏ ퟐ 퐠 퐭ퟐ---------- (1) b) Considering horizontal motion of projectile [Horizontal range ] = [퐻표푟푖푧표푛푡푎푙 푣푒푙표푐푖푡푦] 푥 [Time of flight] x = u cos α x t --------- (2) Velocity with which it strikes the ground (VB): VB = u = initial velocity Chapter No. 7 Circular Motion Page 11
  • 12. VH = u cos α VV = u sin α We have, 퐯 = 퐮 + 퐚퐭 -VV = u sin α – gt VV = -u sin α + gt ퟐ + 푽푽 VB = √ 푽푯 ퟐ--------------- (velocity of hit) Direction of hit: θ = tan-1 ( 푽푽 푽푯 ) ------------ (with the horizontal) Problems: 1. A body is projected from the top of the tower 40 m high and strikes the ground after 10 seconds at a point 400 m from the base of the tower. Determine the velocity and angle of projection. Also determine the maximum height attained by the body above ground level. Dec. 2009 (10 MKS) Ans: u = 60.24 m/s and Hmax = 143.43 m 2. A cricket ball is thrown by a fielder from a height of 2 , at an angle of elevation 300 to the horizontal with an initial velocity of 20 m/s, hits the wickets at a height of 0.5 m from the ground. How far the fielder from the wicket? (May 2004 12 MKS) Ans: t = 2.181 sec, x = 37.77 m 3. A solider fires a bullet at an angle of elevation 300 from his position on the top of hill to strike a target which is 60 m lower than the position of the solider. The initial velocity of the bullet is 75 m/s. Calculate: a) Hmax to which bullet rise above horizontal b) The actual velocity with which it will strike the target c) Total time required by for the flight of the bullet Chapter No. 7 Circular Motion Page 12
  • 13. Ans: Hmax = 131.74 m, t = 9.01 sec, vB = 82.456 m/s, θ = 38.020 4. A projectile fired from the edge of a 150 m high cliff with an initial velocity of 180 m/s at an angle of elevation of 300 with the horizontal. Neglecting air resistance. Find: 1) The greatest elevation above the ground reached by the projectile and 2) Horizontal distance from gun to the point, where the projectile strikes the ground. Ans: Hmax = 563.3 m, t = 19.9 sec, R = 3.012 km Case III) Projectile projected at an angle α from certain height to the downward (say y) c) Considering vertical motion of projectile We have, S = 퐮퐭 + ퟏ ퟐ 퐚퐭ퟐ Here, S = - y = vertically downward displacement u = initial vertical velocity = - u sin α a = acceleration due to gravity = - g = 9.81 m/s2 ∴ − y = ( - u sin α) – ퟏ ퟐ 퐠 퐭ퟐ y = (u sin α) + ퟏ ퟐ 퐠 퐭ퟐ---------- (1) d) Considering horizontal motion of projectile [Horizontal range ] = [퐻표푟푖푧표푛푡푎푙 푣푒푙표푐푖푡푦] 푥 [Time of flight] Chapter No. 7 Circular Motion Page 13
  • 14. x = u cos α x t --------- (2) Velocity with which it strikes the ground (VB): VB = u = initial velocity VH = u cos α VV = u sin α We have, 퐯 = 퐮 + 퐚퐭 -VV = - u sin α – gt VV = u sin α + gt ퟐ + 푽푽 VB = √ 푽푯 ퟐ--------------- (velocity of hit) Direction of hit: θ = tan-1 ( 푽푽 푽푯 ) ------------ (with the horizontal) Problems: 1) A stone is thrown from the top of the tower 30 m high at an angle of 400 with horizontal with velocity 40 kmph. Find when and where the stone will hit a horizontal ground. Also find the velocity and direction of striking. Ans: t = 1.85 sec, x = 15.74 m, Vb = 26.67 m/s, θ = 71.390 Chapter No. 7 Circular Motion Page 14