Chapter 3 Motion in a Plane (two dimension) Part 1 – Physics free study material by TEACHING CARE online tuition and coaching classes
The motion of an object is called two dimensional, if two of the three coordinates are required to specify the position of the object in space changes w.r.t time.
In such a motion, the object moves in a plane. For example, a billiard ball moving over the billiard table, an insect crawling over the floor of a room, earth revolving around the sun etc.
Two special cases of motion in two dimension are 1. Projectile motion 2. Circular motion
Introduction.
A hunter aims his gun and fires a bullet directly towards a monkey sitting on a distant tree. If the monkey remains in his position, he will be safe but at the instant the bullet leaves the barrel of gun, if the monkey drops from the tree, the bullet will hit the monkey because the bullet will not follow the linear path.
The path of motion of a bullet will be parabolic and this motion of bullet is defined as projectile motion.
If the force acting on a particle is oblique with initial velocity then the motion of particle is called projectile motion.
Projectile.
A body which is in flight through the atmosphere but is not being propelled by any fuel is called projectile.
Example: (i) A bomb released from an aeroplane in level flight
(ii) A bullet fired from a gun
 An arrow released from bow
 A Javelin thrown by an athlete
Assumptions of Projectile Motion.
(1) There is no resistance due to air.
 The effect due to curvature of earth is
(3) The effect due to rotation of earth is negligible.
 For all points of the trajectory, the acceleration due to gravity ‘g’ is constant in magnitude and
Principles of Physical Independence of Motions.
(1) The motion of a projectile is a twodimensional motion. So, it can be discussed in two parts. Horizontal motion and vertical motion. These two motions take place independent of each other. This is called the principle of physical independence of motions.
 The velocity of the particle can be resolved into two mutually perpendicular components. Horizontal component and vertical
(3) The horizontal component remains unchanged throughout the flight. The force of gravity continuously affects the vertical component.
 The horizontal motion is a uniform motion and the vertical motion is a uniformly accelerated retarded motion.
Types of Projectile Motion.
(1) Oblique projectile motion (2) Horizontal projectile motion (3) Projectile motion on an inclined plane
Oblique Projectile.
In projectile motion, horizontal component of velocity (u cosq), acceleration (g) and mechanical energy remains constant while, speed, velocity, vertical component of velocity (u sin q), momentum, kinetic energy and potential energy all changes. Velocity, and KE are maximum at the point of projection while minimum (but not zero) at highest point.
 Equation of trajectory : A projectile thrown with velocity u at an angle q with the horizontal. The velocity u can be resolved into two rectangular
v cos q component along X–axis and u sin q component along Y–axis.
For horizontal motion x = u cosq ´ t Þ
t = x
u cosq
…. (i)
For vertical motion
y = (u sinq ) t – 1 gt ^{2}
2
…. (ii)
From equation (i) and (ii)
æ
y = u sinq ç
æ 2 ö








÷ ç ÷


2 2
y = x tanq – 1
gx ^{2}
è u cosq ø ç
cos ÷
2 u^{2} cos ^{2} q
This equation shows that the trajectory of projectile is parabolic because it is similar to equation of parabola
y = ax – bx^{2}
Note : @ Equation of oblique projectile also can be written as
y = x tanq é1 – x ù
(where R = horizontal range =
u^{2} sin 2q )
ëê R úû g
Problem 1. The trajectory of a projectile is represented by y = 3x – gx ^{2} /2 . The angle of projection is
 30^{o} (b) 45^{o} (c) 60^{o} (d) None of these
Solution : (c) By comparing the coefficient of x in given equation with standard equation y = x tanq –
gx ^{2}
2u ^{2} cos ^{2} q
tanq = \q = 60°
Problem 2. The path followed by a body projected along yaxis is given as by y =
initial velocity of projectile will be – (x and y are in m)
3x – (1 / 2)x ^{2} , if g = 10 m/s, then the
(a) 3
m/s
(b)
2 10 m/s
(c)
10 3 m/s
(d) 10
 m/s
Solution : (b) By comparing the coefficient of x^{2} in given equation with standard equation y = x tanq –
g = 1
gx ^{2} .
2u ^{2} cos ^{2} q
2u ^{2} cos ^{2} q 2
Substituting q = 60^{o} we get u = 2 10 m / sec .
Problem 3. The equation of projectile is y = 16x – 5x 2 . The horizontal range is
4
(a) 16 m (b) 8 m (c) 3.2 m (d) 12.8 m
Solution : (d) Standard equation of projectile motion y = x tanq é1 – x ù
ëê R úû
Given equation : y = 16x – 5x 2 or y = 16xé1 – x ù
4 êë 64 / 5 úû
By comparing above equations R = 64
5
=12.8 m.
(2) Displacement of projectile (r) : Let the particle acquires a position P having the coordinates (x, y) just after time t from the instant of projection. The corresponding position vector of the particle at time t is r as shown in the figure.

r
r = xi + yj
….(i)
The horizontal distance covered during time t is given as
x = vx t Þ x = u cosq t
….(ii)
The vertical velocity of the particle at time t is given as
vy = (v0 )y – gt,
….(iii)
Now the vertical displacement y is given as
y = u sinq t – 1 / 2 gt ^{2}
….(iv)
Putting the values of x and y from equation (ii) and equation (iv) in equation (i) we obtain the position vector at any time t as
r ˆ æ
1 _{2} ö ˆ

r = (u cosq ) t i + ç(u sinq ) t – gt ÷ j
è ø
Þ r =
1
_{1}æ ut sinq – 1 / 2gt ^{2} ö
_{1}æ 2u sinq – gt ö
r = u t
and f = tan
(y / x)
= tan
ç ÷ or f = tan ç ÷


ç (u t cosq ) ÷ è 2u cosq ø
Note : @The angle of elevation f of the highest point of the projectile and the angle of projection q are related to each other as
tan f = 1 tanq
2
Problem 4. A body of mass 2 kg has an initial velocity of 3 m/s along OE and it
is subjected to a force of 4 Newton’s in OF direction perpendicular to OE. The distance of the body from O
after 4 seconds will be
(a) 12 m (b) 28 m (c) 20 m (d) 48 m
Solution : (c) Body moves horizontally with constant initial velocity 3 m/s upto 4 seconds \
x = ut = 3 ´ 4 = 12 m
and in perpendicular direction it moves under the effect of constant force with zero initial velocity upto 4 seconds.
\ y = ut + 1 (a) t ^{2}
= 0 + 1 æ F ö t ^{2} = 1 æ 4 ö 4^{2} = 16 m
ç ÷ ç ÷
2 2 è m ø 2 è 2 ø
So its distance from O is given by d = =
\ d = 20 m
Problem 5. A body starts from the origin with an acceleration of 6 m/s^{2} along the xaxis and 8 m/s^{2} along the yaxis. Its distance from the origin after 4 seconds will be [MP PMT 1999]
(a) 56 m (b) 64 m (c) 80 m (d) 128 m
Solution : (c) Displacement along X axis : x = ux t + 1 ax t ^{2} = 1 ´ 6 ´ (4)^{2} = 48 m
2 2
Displacement along Y axis : y = uy t + 1 ay t ^{2} = 1 ´ 8 ´ (4)^{2} = 64 m
2 2
Total distance from the origin = =
= 80 m
(3) Instantaneous velocity v : In projectile motion, vertical component of velocity changes but horizontal component of velocity remains always constant.
Example : When a man jumps over the hurdle leaving behind its skateboard then vertical component of his velocity is changing, but not the horizontal component, which matches with the skateboard velocity.
As a result, the skateboard stays underneath him, allowing him to land on it.
Let v_{i} be the instantaneous velocity of projectile at time t direction of this velocity is along the tangent to the trajectory at point P.
r ˆ Þ
vi = vxi + vy j
vi =
vi = =
Direction of instantaneous velocity tana = vy
= u sinq – gt
or a = tan ^{1} étanq – gt sec q ù
vx u cosq
ëê u úû
 Change in velocity : Initial velocity (at projection point) u_{i} = u cosq ˆi + u sinq ˆj
Final velocity (at highest point) u _{f}
= u cosq ˆi + 0 ˆj
(i) Change in velocity (Between projection point and highest point) Du = u _{f} – u_{i} = –u sinq ˆj
When body reaches the ground after completing its motion then final velocity u _{f}
= u cosq ˆi – u sinq ˆj
(ii) Change in velocity (Between complete projectile motion) Du = uf – ui = 2u sinq ˆi
Problem 6. In a projectile motion, velocity at maximum height is [AIEEE 2002]
 u cosq
2
u cos q
u sinq 2
 None of these
Solution : (b) In a projectile motion at maximum height body possess only horizontal component of velocity i.e. u cosq.
Problem 7. A body is thrown at angle 30^{o} to the horizontal with the velocity of 30 m/s. After 1 sec, its velocity will be (in
m/s) (g = 10 m/s^{2})
 10
700 (c)
100 (d)
Solution : (a) From the formula of instantaneous velocity v =
v = = 10 7 m / s
Problem 8. A projectile is fired at 30^{o} to the horizontal. The vertical component of its velocity is 80 ms^{–1}. Its time of flight is
 T. What will be the velocity of the projectile at t = T/2
(a) 80 ms^{–1} (b) 80
ms^{–1} (c) (80/
) ms^{–1} (d) 40 ms^{–1}
Solution : (b) At half of the time of flight, the position of the projectile will be at the highest point of the parabola and at that position particle possess horizontal component of velocity only.
Given uvertical
= u sinq = 80 Þ u =
80 = 160 m / s
sin 30 ^{o}
\ uhorizontal = u cosq = 160 cos 30 ^{o} = 80
 m / s.
Problem 9. A particle is projected from point O with velocity u in a direction making an angle a
with the horizontal. At
any instant its position is at point P at right angles to the initial direction of projection. Its velocity at point P is
 u tana
 u cota
 u coseca
 u seca
Solution : (b) Horizontal velocity at point ‘ O‘ = u cosa
Horizontal velocity at point ‘ P‘ = v sin a
In projectile motion horizontal component of velocity remains constant throughout the motion
\ v sin a = u cos a Þ v = u cot a
Problem 10. A particle P is projected with velocity u_{1} at an angle of 30^{o} with the horizontal. Another particle Q is thrown vertically upwards with velocity u_{2} from a point vertically below the highest point of path of P. The necessary condition for the two particles to collide at the highest point is
 u1 = u2
 u1 = 2u2
 u1 = u2
2
 u1 = 4u2
Solution : (b) Both particle collide at the highest point it means the vertical distance travelled by both the particle will be equal, i.e. the vertical component of velocity of both particle will be equal
u_{1} sin 30° = u_{2} Þ
u1 = u
2 2
\ u1 = 2u2
Problem 11. Two seconds after projection a projectile is travelling in a direction inclined at 30^{o} to the horizontal after one more sec, it is travelling horizontally, the magnitude and direction of its velocity are [RPET 1999]
(a)
2 20 m/sec, 60 ^{o}
(b)
20 3 m/sec, 60 ^{o}
(c) 6
m/sec, 30 ^{o}
(d)
40 6 m/sec, 30 ^{o}
Solution : (b) Let in 2 sec body reaches upto point A and after one more sec upto point B.
Total time of ascent for a body is given 3 sec i.e.
t = u sinq = 3
g
\ u sinq = 10 ´ 3 = 30 …..(i)
Horizontal component of velocity remains always constant
u cosq = v cos 30° …..(ii)
For vertical upward motion between point O and A
v sin 30 ^{o} = u sinq – g ´ 2
v sin 30 ^{o} = 30 – 20
\ v = 20 m / s.
[Using v = u – g t]
[As u sin q = 30]
Substituting this value in equation (ii) From equation (i) and (iii) u = 20
u cosq = 20 cos 30 ^{o} = 10
and q = 60°
…..(iii)
Problem 12. A body is projected up a smooth inclined plane (length = 20
2 m ) with velocity u from the point M as shown
in the figure. The angle of inclination is 45^{o} and the top is connected to a well of diameter 40 m. If the body just manages to cross the well, what is the value of v
 40 ms ^{–}^{1}
(b) 40 2 ms ^{–}^{1}
(c) 20 ms ^{–}^{1}
(d) 20 2 ms ^{–}^{1}
Solution : (d) At point N angle of projection of the body will be 45°. Let velocity of projection at this point is v.
If the body just manages to cross the well then Range = Diameter of well
v ^{2} sin 2q
g
v^{2} = 400
= 40
Þ
[As q = 45°]
v = 20 m / s
But we have to calculate the velocity (u) of the body at point M.
For motion along the inclined plane (from M to N) Final velocity (v) = 20 m/s,
acceleration (a) = – g sina = – g sin 45^{o}, distance of inclined plane (s) = 20 m
(20)^{2} = u ^{2} – 2
g .20
[Using v^{2} = u^{2} + 2as]
u ^{2} = 20^{2} + 400
Þ u = 20
2 m / s.
Problem 13. A projectile is fired with velocity u making angle q with the horizontal. What is the change in velocity when it is at the highest point
 u cosq (b) u (c) u sinq (d) (u cosq – u)
Solution : (c) Since horizontal component of velocity remain always constant therefore only vertical component of velocity changes.
Initially vertical component u sinq
Finally it becomes zero. So change in velocity = u sinq
(5) Change in momentum : Simply by the multiplication of mass in the above expression of velocity (Article4).
 Change in momentum (Between projection point and highest point)
Dp = p f – pi = – mu sinq ˆj
(ii) Change in momentum (For the complete projectile motion) Dp = p f – pi = – 2mu sinq ˆj
 Angular momentum : Angular momentum of projectile at highest point of trajectory about the point of projection is given by
L = mvr
éHere r = H = u2 sin 2 q ù
êê 2g ú

\ L = m u cosq
u^{2} sin ^{2} q

2g
= m u^{3} cosq sin ^{2} q
2g
Problem 14. A body of mass 0.5 kg is projected under gravity with a speed of 98 m/s at an angle of 30^{o} with the horizontal. The change in momentum (in magnitude) of the body is [MP PET 1997]
(a) 24.5 N–s (b) 49.0 N–s (c) 98.0 N–s (d) 50.0 N–s
Solution : (b) Change in momentum between complete projectile motion = 2mu sinq = 2 ´ 0.5 ´ 98 ´ sin 30° = 49 N–s.
Problem 15. A particle of mass 100 g is fired with a velocity 20 m sec^{–1} making an angle of 30^{o} with the horizontal. When it rises to the highest point of its path then the change in its momentum is
(a)
3kg m sec ^{–}^{1}
 1/2 kg m sec^{–1} (c)
2 kg m sec ^{–}^{1}
 1 kg m sec^{–1}
Solution : (d) Horizontal momentum remains always constant
So change in vertical momentum (D p ) = Final vertical momentum – Initial vertical momentum = 0 – mu sinq
 DP = 0.1 ´ 20 ´ sin 30 ^{o} = 1kg m / sec .
Problem 16. Two equal masses (m) are projected at the same angle (q) from two points separated by their range with equal velocities (v). The momentum at the point of their collision is
 Zero (b) 2 mv cosq (c) – 2 mv cosq (d) None of these
Solution : (a) Both masses will collide at the highest point of their trajectory with equal and opposite momentum. So net momentum of the system will be zero.
Problem 17. A particle of mass m is projected with velocity v making an angle of 45^{o} with the horizontal. The magnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where g = acceleration due to gravity) [MP PMT 1994; UPSEAT 2000; MP PET 2001]
(a) Zero (b) mv^{3}/ (4
2g)
 mv^{3}/ (
2g)
 mv^{2}/2g
Solution : (b)
L = m u ^{3} cosq sin ^{2} q =
2g
mv ^{3}
[As q = 45^{o}]
Problem 18. A body is projected from the ground with some angle to the horizontal. What happens to the angular momentum about the initial position in this motion [AIIMS 2000]
 Decreases (b) Increases
(c) Remains same (d) First increases and then decreases
Solution : (b)
Problem 19. In case of a projectile, where is the angular momentum minimum
 At the starting point
 At the highest point
 On return to the ground
 At some location other than those mentioned above
Solution : (a)
(7) Time of flight : The total time taken by the projectile to go up and come down to the same level from which it was projected is called time of flight.
For vertical upward motion 0 = u sin q – gt Þ t = (u sin q/g)
Now as time taken to go up is equal to the time taken to come down so
Time of flight T = 2t = 2u sinq
g
(i) Time of flight can also be expressed as :
T = 2.uy
g
(where u_{y}
is the vertical component of initial velocity).
 For complementary angles of projection q and 90^{o} – q
(a) Ratio of time of flight = T1
T2
= 2u sinq / g
2u sin(90 – q ) / g
= tanq
Þ T1 = tanq
T2
(b) Multiplication of time of flight = T1T2
= 2u sinq 2u cosq
Þ T1T2 = 2R
g g g
(iii) If t_{1} is the time taken by projectile to rise upto point p and t_{2} is the time taken in falling from point p to
ground level then
t1 + t 2 = 2u sinq

g(t1 + t 2 )
= time of flight
or u sinq =
2
and height of the point p is given by h = u sinq t
 1gt ^{2}
1 2 1
h = g (t1 + t 2 ) t
 1gt ^{2}
2
by solving h = g t1t 2
2
1 2 1
(iv) If B and C are at the same level on trajectory and the time difference between these two points is t_{1}, similarly A and D are also at the same level and the time difference between these two positions is t_{2} then
t ^{2} – t ^{2} = 8h
2 1 g
Problem 20. For a given velocity, a projectile has the same range R for two angles of projection if t_{1} and t_{2} are the times of flight in the two cases then [KCET 2003]
(a)
t1t 2 µ R ^{2}
(b)
t1t 2 µ R
t1t 2 µ 1
R
t1t 2 µ 1
R 2
Solution : (b) As we know for complementary angles t1 t 2 = 2R
g
\ t1 t 2 µ R .
Problem 21. A body is thrown with a velocity of 9.8 m/s making an angle of 30^{o} with the horizontal. It will hit the ground after a time [JIPMER 2001, 2002; KCET (Engg.) 2001]
(a) 1.5 s (b) 1 s (c) 3 s (d) 2 s
Solution : (b)
T = 2u sinq
g
= 2 ´ 9.8 ´ sin 30 ^{o} = 1sec
9.8
Problem 22. Two particles are separated at a horizontal distance x as shown in figure. They are projected at the same time as shown in figure with different initial speed. The time after which the horizontal distance between the particles become zero is [CBSE PMT 1999]
(a)
u /2x
(b)
x/u
(c)
2u/x
(d)
u/x
Solution : (b) Let
x_{1} and x_{2} are the horizontal distances travelled by particle A and B respectively in time t.
x1 = u . cos 30° ´ t
…..(i) and
x _{2} = u cos 60 ^{o} ´ t
……(ii)
x1 + x 2 = u . cos 30 ^{o} ´ t + u cos 60 ^{o} ´ t = ut
Þ x = ut
\ t = x / u
Problem 23. A particle is projected from a point O with a velocity u in a direction making an angle a upward with the horizontal. After some time at point P it is moving at right angle with its initial direction of projection. The time of flight from O to P is
(a)
u sina g
(b)
u coseca
g
(c)
u tana
g
(d)
u sec a
g
Solution : (b) When body projected with initial velocity u by making angle a with the horizontal. Then after time t, (at point P) it’s direction is perpendicular to u .
Magnitude of velocity at point P is given by v = u cot a. (from sample problem no. 9) For vertical motion : Initial velocity (at point O) = u sina
Final velocity (at point P) = –v cosa = –u cot a cosa
Time of flight (from point O to P) = t
Applying first equation of motion v = u – g t
 u cot a cos a = u sin a – g t
\ t = u sin a + u cot a cos a
= u [sin ^{2} a + cos ^{2} a ] = u coseca
g g sin a g
Problem 24. A ball is projected upwards from the top of tower with a velocity 50 ms^{–1} making angle 30^{o} with the horizontal. The height of the tower is 70 m. After how many seconds from the instant of throwing will the ball reach the ground
(a) 2.33 sec (b) 5.33 sec (c) 6.33 sec (d) 9.33 sec
Solution : (c) Formula for calculation of time to reach the body on the ground from the tower of height ‘h’ (If it is thrown
vertically up with velocity u) is given by t = u é +
1 + 2gh ù

g ê1
ú
u^{2} úû
So we can resolve the given velocity in vertical direction and can apply the above formula.
Initial vertical component of velocity u sinq = 50 sin 30 = 25 m / s.
\ t = 25 é +
1 + 2 ´ 9.8 ´ 70 ù
= 6.33 sec.

9.8 ê1
(25)^{2} ú

Problem 25. If for a given angle of projection, the horizontal range is doubled, the time of flight becomes
Solution : (c)
(a) 4 times (b) 2 times (c) times (d) 1 /
R = u ^{2} sin 2q and T = 2u sinq
times
g g
\ R µ u^{2} and T µ u (If q and g are constant).
In the given condition to make range double, velocity must be increased upto times that of previous value.
So automatically time of flight will becomes times.
Problem 26. A particle is thrown with velocity u at an angle a from the horizontal. Another particle is thrown with the same velocity at an angle a from the vertical. The ratio of times of flight of two particles will be
(a) tan 2 a : 1 (b) cot 2 a : 1 (c) tan a : 1 (d) cot a : 1
Solution : (c) For first particles angle of projection from the horizontal is a. So
T1 = 2u sina
g
For second particle angle of projection from the vertical is a. it mean from the horizontal is (90 – a).
\ T = 2u sin(90 – a) = 2u cosa . So ratio of time of flight T1 = tana .
^{2} g g T2
Problem 27. The friction of the air causes vertical retardation equal to one tenth of the acceleration due to gravity (Take
g = 10 ms^{–2}). The time of flight will be decreased by
(a) 0% (b) 1% (c) 9% (d) 11%
q T g
g + g
Solution : (c)
T = 2u sin
\ 1 = 2 = 10
= 11
g T2 g1
g 10
Fractional decrease in time of flight = T1 – T2
T1
= 1
11
Percentage decrease = 9%
(8) Horizontal range : It is the horizontal distance travelled by a body during the time of flight. So by using second equation of motion
R = u cosq ´ T =
u^{2} sin 2q
u cosq ´ (2u sinq / g) = u2 sin 2q
g
R = g
(i) Range of projectile can also be expressed as :
R = u cosq × T = u cosq 2u sinq
g
= 2u cosq u sinq
g
= 2ux uy
g
\ R = 2ux uy
g
(where u_{x}
and u_{y}
are the horizontal and vertical component of initial velocity)
 If angle of projection is changed from q to
q ¢ = (90 – q) then range remains unchanged.
R‘ = u^{2} sin 2q ‘ = u^{2} sin[2(90^{o} – q )] = u^{2} sin 2q = R
g g g
So a projectile has same range at angles of projection q and (90 – q), though time of flight, maximum height and trajectories are different.
These angles q and 90^{o} – q are called complementary angles of projection and for complementary angles of
projection ratio of range
R1 =
R2
u^{2} sin 2q / g
u^{2} sin[2(90^{o} – q )] / g
= 1 Þ
R1 = 1
R2
(iii) For angle of projection q_{1} = (45 – a) and q_{2} = (45 + a), range will be same and equal to u^{2} cos 2a/g.
q_{1} and q_{2} are also the complementary angles.
(iv) Maximum range : For range to be maximum
dR d
éu ^{2} sin 2q ù
dq = 0 Þ
dq ê g
ú = 0
úû

Þ cos 2q = 0 i.e. 2q = 90^{o} Þ q = 45^{o} and R_{max} = (u^{2}/g)
i.e., a projectile will have maximum range when it is projected at an angle maximum range will be (u^{2}/g).
When the range is maximum, the height H reached by the projectile
of 45^{o} to the horizontal and the
H = u ^{2} sin ^{2} q
= u ^{2} sin ^{2} 45 = u ^{2}
= Rmax
2g 2g 4 g 4
i.e., if a person can throw a projectile to a maximum distance R , The maximum height to which it will rise is æ Rmax ö .

max
(v) Relation between horizontal range and maximum height :
R = u^{2} sin 2q
g
and
ç ÷
è ø
H = u^{2} sin ^{2} q
2g
\ R
H
= u^{2} sin 2q / g u^{2} sin ^{2} q / 2g
= 4 cotq
Þ R = 4 H cotq
(vi) If in case of projectile motion range R is n times the maximum height H
i.e. R = nH Þ
u ^{2} sin 2q = n u ^{2} sin ^{2} q Þ
tanq = [4 / n]
or q = tan ^{1}[4 / n]
g
The angle of projection is given by q
2g
= tan ^{1}[4 / n]
Note : @ If R = H then q = tan ^{1}(4) or q = 76^{o} .
If R = 4H then q = tan ^{1}(1) or q = 45^{o} .
Problem 28. A boy playing on the roof of a 10m high building throws a ball with a speed of 10 m/s at an angle of 30^{o} with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground
(g = 10 m/s^{2}, sin 30^{o} =
1 , cos 30^{o} =
2
3 ) [AIEEE 2003]
2
(a) 8.66 m (b) 5.20 m (c) 4.33 m (d) 2.60 m
Solution : (a) Simply we have to calculate the range of projectile
R = u ^{2} sin 2q
g
= (10)^{2} sin(2 ´ 30°)
10
R = 5 = 8.66 meter
Problem 29. Which of the following sets of factors will affect the horizontal distance covered by an athlete in a long–jump event [AMU (Engg.) 2001]
 Speed before he jumps and his weight
 The direction in which he leaps and the initial speed
 The force with which he pushes the ground and his speed
 The direction in which he leaps and the weight
Solution : (b) Because range = (Velocity of projection)^{2} ´ sin 2(Angle of projection)
g
Problem 30. For a projectile, the ratio of maximum height reached to the square of flight time is (g = 10 ms^{–2})
[EAMCET (Med.) 2000]
(a) 5 : 4 (b) 5 : 2 (c) 5 : 1 (d) 10 : 1
Solution : (a)
H = u^{2} sin ^{2} q and T = 2u sinq
\ H = u ^{2} sin ^{2} q / 2g
= g = 10 = 5
2g g
T ^{2} 4 u ^{2} sin ^{2} q / g ^{2}
8 8 4
Problem 31. A cricketer can throw a ball to a maximum horizontal distance of 100 m. The speed with which he throws the ball is (to the nearest integer) [Kerala (Med.) 2002]
(a) 30 ms^{–1} (b) 42 ms^{–1} (c) 32 ms^{–1} (d) 35 ms^{–1}
Solution : (c)
Rmax
= u2 = 100 (when q = 45° ) \ u =
g
= 31.62m / s.
Problem 32. If two bodies are projected at 30^{o} and 60^{o} respectively, with the same velocity, then [CBSE PMT 2000; JIPMER 2002]
(a) Their ranges are same (b) Their heights are same
(c) Their times of flight are same (d) All of these
Solution : (a) Because these are complementary angles.
Problem 33. Figure shows four paths for a kicked football. Ignoring the effects of air on the flight, rank the paths according to initial horizontal velocity component, highest first [AMU (Med.) 2000]
(a) 1, 2, 3, 4 (b) 2, 3, 4, 1 (c) 3, 4, 1, 2 (d) 4, 3, 2, 1
Solution : (d) Range µ horizontal component of velocity. Graph 4 shows maximum range, so football possess maximum horizontal velocity in this case.
Problem 34. Four bodies P, Q, R and S are projected with equal velocities having angles of projection 15^{o}, 30^{o}, 45^{o} and 60^{o} with the horizontal respectively. The body having shortest range is [EAMCET (Engg.) 2000]
(a) P (b) Q (c) R (d) S Solution : (a) Range of projectile will be minimum for that angle which is farthest from 45°.
Problem 35.  A particle covers 50 m distance when projected with an initial speed.  On the same surface it will cover a 
distance, when projected with double the initial speed
(a) 100 m (b) 150 m (c) 200 m 
[RPMT 2000]
(d) 250 m 
u^{2} sin 2q
R æ u ö ^{2}
æ 2u ö ^{2}
Solution : (c) R =
\ R µ u ^{2} so ^{ } ^{2} = ç 2 ÷ = ç
÷ Þ R_{2} = 4 R_{1} = 4 ´ 50 = 200 m
g R_{1}
è u1 ø
è u ø
Problem 36. A bullet is fired from a canon with velocity 500 m/s. If the angle of projection is 15^{o} and g = 10 m/s^{2}. Then the range is [CPMT 1997]
(a) 25 × 10^{3} m (b) 12.5 × 10^{3} m (c) 50 × 10^{2} m (d) 25 × 10^{2} m
Solution : (b)
Range (R) = u 2 sin 2q
g
= (500)2 sin(2 ´ 15) = 12500 m
10
= 12.5 ´ 10^{3} m
Problem 37. A projectile thrown with a speed v at an angle q has a range R on the surface of earth. For same v and q, its range on the surface of moon will be
(a) R/6 (b) 6 R (c) R/36 (d) 36 R
Solution : (b)
R = u^{2} sin 2q
g
\ R µ 1 / g
RMoon g Earth = 6 é 1 ù
R = g êQ g Moon = 6 g Earth ú
Earth
Moon ë û
\RMoon = 6 REarth = 6R
Problem 38. A projectile is thrown into space so as to have maximum horizontal range R. Taking the point of projection as origin, the coordinates of the point where the speed of the particle is minimum are
(a) (R, R) (b)
æ R, R ö
(c)
æ R , R ö
(d)
æ R, R ö




ç ÷ ç ÷ ç ÷
è ø è ø è ø
Solution : (c) For maximum horizontal Range q = 45°
From R = 4 H cotq = 4H [As q = 45^{o}, for maximum range.]
Speed of the particle will be minimum at the highest point of parabola.
So the coordinate of the highest point will be (R/2, R/4)
Problem 39. The speed of a projectile at the highest point becomes 1
times its initial speed. The horizontal range of the
projectile will be
(a)
u2 (b)
g
u2 (c)
2g
u2 (d) u2
3g 4 g
Solution : (a) Velocity at the highest point is given by u cosq = u
(given) \ q = 45^{o}
Horizontal range R = u 2 sin 2q
= u ^{2} sin(2 ´ 45 ^{o} ) = u ^{2}
g g g
Problem 40. A large number of bullets are fired in all directions with same speed u. What is the maximum area on the ground on which these bullets will spread
(a)
p u 2
g
(b)
p u 4
g 2
(c)
p 2 u 4
g 2
(d)
p 2 u 2
g 2
Solution : (b) The maximum area will be equal to area of the circle with radius equal to the maximum range of projectile
Maximum area pr ^{2} = p (Rmax )2
æ 2 ö ^{2}

= p ç ÷
= p u4

[As r = Rmax = u ^{2} / g
for q = 45^{o}]

ç ÷
è ø
Problem 41. A projectile is projected with initial velocity (6ˆi + 8ˆj)m / sec. If g = 10 ms^{–}^{2}, then horizontal range is
(a) 4.8 metre (b) 9.6 metre (c) 19.2 metre (d) 14.0 metre
Solution : (b) Initial velocity = (6ˆi + 8 Jˆ)m / s
(given)
Magnitude of velocity of projection u = = = 10 m/s
Angle of projection tanq = uy
= 8 = 4
\ sinq = 4 and cosq = 3
u_{x} 6 3
2 2
5 5
(10)^{2} ´ 2 ´ 4 ´ 3
Now horizontal range R = u sin 2q
g
= u 2 sinq cosq
g
= 5 5
10
= 9.6 meter
Problem 42. A projectile thrown with an initial speed u and angle of projection 15^{o} to the horizontal has a range R. If the same projectile is thrown at an angle of 45^{o} to the horizontal with speed 2u, its range will be
(a) 12 R (b) 3 R (c) 8 R (d) 4 R
Solution : (c)
R = u^{2} sin 2q
g
\ R µ u^{2} sin 2q
R æ u ö ^{2} æ sin 2q ö
æ 2u ö ^{2} æ sin 90 ^{o} ö
2 = ç 2 ÷
ç 2 ÷ Þ R_{2} = R_{1} ç ÷ ç ÷ = 8R_{1}
R1 è u1 ø
è sin 2q_{1} ø
è u ø
ç sin 30 ^{o} ÷
è ø
Problem 43. The velocity at the maximum height of a projectile is half of its initial velocity of projection u. Its range on the horizontal plane is [MP PET 1993]
(a)
3u^{2} / 2g
(b)
u^{2} / 3g
(c)
3u^{2} / 2g
(d)
3u^{2} / g
Solution : (a) If the velocity of projection is u then at the highest point body posses only u cosq
u cosq = u
2
(given)
\ q = 60°
Now R = u^{2} sin(2 ´ 60°) =
g 2 g
Problem 44. A projectile is thrown from a point in a horizontal place such that its horizontal and vertical velocity component are 9.8 m/s and 19.6 m/s respectively. Its horizontal range is
(a) 4.9 m (b) 9.8 m (c) 19.6 m (d) 39.2 m
Solution : (d) We know R = 2ux uy
g
= 2 ´ 9.8 ´ 19.6
9.8
= 39.2 m
Where ux = horizontal component of initial velocity, uy = vertical component of initial velocity.
Problem 45. A particle is projected with a velocity v such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where g is acceleration due to gravity) [BHU 1984]
(a)
4v^{2}
5g
(b)
4g (c)
5v ^{2}
v2 (d)
g
4v^{2}
Solution : (a) We know R = 4 H cotq
2H = 4 H cotq
Þ cot q = 1
2
; sinq = 2 ; cosq = 1
5
[As R = 2H given]
Range = u ^{2} . 2. sinq . cosq =
g g
= 4u^{2}
5g
Problem 46. The range R of projectile is same when its maximum heights are h_{1} and h_{2}. What is the relation between R and
h_{1} and h_{2} [EAMCET (Med.) 2000]
 R =
(b)
R = (c)
R = 2
(d)
R = 4
Solution : (d) For equal ranges body should be projected with angle q or (90 ^{o} – q ) from the horizontal.
And for these angles : h_{1}
= u^{2} sin ^{2} q
2g
and h2
= u ^{2} cos ^{2} q
2g
by multiplication of both height : h1 h2 =
u ^{2} sin ^{2}q cos ^{2}q 1 æ u ^{2} sin 2q ö 2





2
Þ 16h1 h2 = R ^{2} Þ R = 4
4 g 16 ç ÷
Problem 47. A grasshopper can jump maximum distance 1.6 m. It spends negligible time on the ground. How far can it go in 10 seconds
(a)
5 2 m
 10 2 m
20 2 m
40 2 m
Solution : (c) Horizontal distance travelled by grasshopper will be maximum for q = 45°
Rmax
= u 2
g
= 1.6 m
Þ u = 4 m / s.
Horizontal component of velocity of grasshopper u cosq = 4 cos 45 = 2 m / s
Total distance covered by it in10 sec.
S = u cosq ´ t = 2
´ 10
= 20 2 m
Problem 48. A projectile is thrown with an initial velocity of v = aˆi + bˆj,
height reached by it then
if the range of projectile is double the maximum
 a = 2b (b) b = a (c) b = 2a (d) b = 4a
Solution : (c) Angle of projection q = tan^{–}^{1} vy = tan^{–}^{1} b
\ tanq = b
…(i)
From formula
vx
R = 4 H cot q = 2H
a
Þ cot q = 1
2
a
\ tanq = 2
…(ii) [As R = 2H given]
From equation (i) and (ii) b = 2a