Chapter 4 Motion in a Plane (two dimension) Part 2 – Physics free study material by TEACHING CARE online tuition and coaching classes
(9) Maximum height : It is the maximum height from the point of projection, a projectile can reach. So, by using v ^{2} = u^{2} + 2as
0 = (u sinq )^{2} – 2gH
u^{2} sin ^{2} q
H = 2g
(i) Maximum height can also be expressed as
u2
H = ^{y}
2g
(where uy
is the vertical component of initial velocity).
H max
= u 2
2g
(when sin^{2}q = max = 1 i.e., q = 90^{o})
i.e., for maximum height body should be projected vertically upward. So it falls back to the point of projection after reaching the maximum height.
 For complementary angles of projection q and 90^{o} – q
Ratio of maximum height =
H1 =
H 2
u^{2} sin ^{2} q / 2g
u^{2} sin ^{2} (90^{o} – q )2g
= sin^{2} q
cos^{2} q
= tan ^{2} q
\ H1
H 2
= tan ^{2} q
Problem 49. A cricketer can throw a ball to a maximum horizontal distance of 100 m. With the same effort, he throws the ball vertically upwards. The maximum height attained by the ball is [UPSEAT 2002]
(a) 100 m (b) 80 m (c) 60 m (d) 50 m
Solution : (d)
Rmax
= u2 = 100 m g
(when q = 45° )
\u ^{2} = 100 ´ 10 = 1000
H max
= u 2
2g
= 1000
2 ´ 10
= 50 metre.
(when q = 90° )
Problem 50. A ball thrown by one player reaches the other in 2 sec. the maximum height attained by the ball above the point of projection will be about [Pb. PMT 2002]
(a) 10 m (b) 7.5 m (c) 5 m (d) 2.5 m
Solution : (c)
T = 2u sinq
g
= 2 sec
(given)
\u sinq = 10
Now
H = u ^{2} sin ^{2} q
2g
= (10)^{2}
2 ´ 10
= 5 m.
Problem 51. Two stones are projected with the same magnitude of velocity, but making different angles with horizontal. The angle of projection of one is p/3 and its maximum height is Y, the maximum height attained by the other stone with as p/6 angle of projection is [J & K CET 2000]
 Y (b) 2 Y (c) 3 Y (d) Y
3
Solution : (d) When two stones are projected with same velocity then for complementary angles q and (90^{o} –q )
Ratio of maximum heights :
H1 = tan ^{2} q
H 2
= tan ^{2} p
3
= 3 Þ
H = H1 = Y
^{2} 3 3
Problem 52. If the initial velocity of a projectile be doubled. Keeping the angle of projection same, the maximum height reached by it will
 Remain the same (b) Be doubled (c) Be quadrupled (d) Be halved
Solution : (c)
H = u^{2} sin 2q
2g
\H µ u^{2}
[As q = constant]
If initial velocity of a projectile be doubled then H will becomes 4 times.
Problem 53. Pankaj and Sudhir are playing with two different balls of masses m and 2m respectively. If Pankaj throws his ball vertically up and Sudhir at an angle q, both of them stay in our view for the same period. The height attained by the two balls are in the ratio
(a) 2 : 1 (b) 1 : 1 (c) 1 : cosq (d) 1 : secq

Solution : (b) Time of flight for the ball thrown by Pankaj T = 2u1
g
Time of flight for the ball thrown by Sudhir T
= 2u2 sin(90 ^{o} – q ) = 2u2 cosq
2
2u 2u
g g
cosq
According to problem T1 = T2
Þ 1 = 2
Þ u_{1} = u_{2} cosq
g g
u 2 Short Trick :
Height of the ball thrown by Pankaj H1 = ^{1}
2g
Maximum height H µ T^{2}
H æ T ö ^{2}
u ^{2} sin ^{2} (90 ^{o} – q )
u ^{2} cos ^{2} q
1 = ç 1 ÷
Height of the ball thrown by Sudhir H _{2} =^{ } ^{2} = ^{ } ^{2} H 2 è T2 ø
H u ^{2} / 2g
2g 2g
\ H1 = 1
H
(As T_{1} = T_{2})
\ 1 = 1
= 1 [As u = u
cosq ] ^{2}

H 2 u ^{2} cos ^{2} q / 2g ^{1} ^{2}
Problem 54. A boy aims a gun at a bird from a point, at a horizontal distance of 100 m. If the gun can impart a velocity of 500 ms^{–1} to the bullet. At what height above the bird must he aim his gun in order to hit it (take g = 10 ms^{–2})
[CPMT 1996]
(a) 20 cm (b) 10 cm (c) 50 cm (d) 100 cm
Solution : (a) Time taken by bullet to travel a horizontal distance of 100 m is given by t = 100 = 1 sec
500 5
In this time the bullet also moves downward due to gravity its vertical displacement
1 _{2} 1 æ 1 ö ^{2}
h = 2 g t = 2 ´ 10 ´ ç 5 ÷
= 1 / 5 m = 20 cm
è ø
So bullet should be fired aiming 20 cm above the bird to hit it.
Problem 55. The maximum horizontal range of a projectile is 400 m. The maximum height attained by it will be (a) 100 m (b) 200 m (c) 400 m (d) 800 m
Solution : (a)
R_{max} = 400 m [when q = 45° ]
So from the Relation R = 4 H cotq Þ
400 = 4 H cot 45° Þ
H = 100 m.
Problem 56. Two bodies are projected with the same velocity. If one is projected at an angle of 30^{o} and the other at an angle of 60^{o} to the horizontal, the ratio of the maximum heights reached is
[EAMCET (Med.) 1995; Pb. PMT 2000; AIIMS 2001]
(a) 3 : 1 (b) 1 : 3 (c) 1 : 2 (d) 2 : 1
Solution : (b)
H1 = sin ^{2} q _{1}
H 2 sin ^{2} q _{2}
= sin ^{2} 30 ^{o} = 1
sin ^{2} 60 ^{o} 3
Problem 57. If time of flight of a projectile is 10 seconds. Range is 500 m. The maximum height attained by it will be
[RPMT 1997; RPET 1998]
(a) 125 m (b) 50 m (c) 100 m (d) 150 m
Solution : (a)
T = 2u sinq
g
= 10 sec Þ
u sinq = 50
so H = u^{2} sin ^{2} q
2g
= (50)2 = 125 m .
2 ´ 10
Problem 58. A man can throw a stone 80 m. The maximum height to which he can raise the stone is
(a) 10 m (b) 15 m (c) 30 m (d) 40 m
Solution : (d) The problem is different from problem no. (54). In that problem for a given angle of projection range was given and we had find maximum height for that angle.
But in this problem angle of projection can vary, R
max
= u2 = 80 m g
[for q = 45° ]
u ^{2} sin ^{2} 90 ^{o}
u ^{2} 1 æ u ^{2} ö



But height can be maximum when body projected vertically up H max = = = ç ÷ = 40 m
2g 2g ç ÷
Problem 59. A ball is thrown at different angles with the same speed u and from the same points and it has same range in both the cases. If y_{1} and y_{2} be the heights attained in the two cases, then y1 + y2 =
(a)
u2 (b)
g
2u^{2}
g
(c)
u2 (d) u2
2g 4 g
Solution : (c) Same ranges can be obtained for complementary angles i.e. q and 90^{o} – q
y = u 2 sin 2 q and y
= u ^{2} cos ^{2} q
\ y + y
= u^{2} sin ^{2} q + u^{2} cos ^{2} q
= u 2
^{1} 2g
^{2} 2g
1 2 2g
2g 2g
(10) Projectile passing through two different points on same height at time t_{1} and t_{2} : If the particle
passes two points situated at equal height y at
t = t_{1} and t = t_{2}, then
(i) Height (y):
y = (u sinq )t_{1} – 1 gt ^{2}
…..(i)
2 1
and
y = (u sinq )t _{2} – 1 gt ^{2}
…..(ii)
2 2
Comparing equation (i) with equation (ii)
u sinq = g(t1 + t 2 )
2
Substituting this value in equation (i)
y = gæ t_{1} + t _{2} ö t
 1gt ^{2}
Þ y = gt1t2
ç ÷ _{1} _{1}

è ø 2 2
(ii) Time (t_{1} and t_{2}):
2u sinq
y = u sinq t – 1 gt ^{2}
2
2y
é

u sinq
æ 2gy ö 2 ù
t ^{2} –
g t + g = 0
Þ t =
ê1 ±
g
ç ÷ ú
ç u sinq ÷ ú
é
u sinq
æ 2gy ö 2 ù
êë

é

u sinq
è ø úû

æ 2gy ö 2 ù

t1 =
ê1 +
g
1 – ç ÷
ç u sinq ÷
ú and
t 2 =
ê1 –
g
ç ÷ ú
ç u sinq ÷ ú
ëê è
ø úû
ëê è
ø úû

 Motion of a projectile as observed from another projectile : Suppose two balls A and B are projected simultaneously from the origin, with initial velocities u_{1} and u_{2} at angle q_{1} and q_{2}, respectively with the
The instantaneous positions of the two balls are given by
Ball A : x_{1}
= (u_{1}
cosq_{1})t
y1 = (u1
sinq _{1}
) t – 1 gt ^{2}
2
Ball B : x_{2} = (u_{2} cosq_{2})t
y2 = (u2 sinq _{2} ) t –
1 gt ^{2}
2
The position of the ball A with respect to ball B is given by
x = x1 – x 2 = (u1 cosq_{1} – u2 cosq _{2} ) t
y = y1 – y2 = (u1 sinq_{1} – u2 sinq _{2} )t
y æ u_{1} sinq_{1} – u_{2} sinq _{2} ö
Now
x = ç u cosq
 u cosq
÷ = constant
è 1 1 2 2 ø
Thus motion of a projectile relative to another projectile is a straight line.
 Energy of projectile : When a projectile moves upward its kinetic energy decreases, potential energy increases but the total energy always remain
If a body is projected with initial kinetic energy K(=1/2 mu^{2}), with angle of projection q with the horizontal then at the highest point of trajectory
(i) Kinetic energy
= 1 m(u cosq )^{2}
2
= 1 mu^{2}
2
cos ^{2} q
\ K‘ = K cos ^{2} q
 Potential energy = mgH
= mg
u ^{2} sin ^{2} q
= 1 mu ^{2}
sin ^{2} q
æ

ç As
H = u^{2} sin ^{2} q ö
2g 2
ç 2g ÷


 Total energy = Kinetic energy + Potential energy = 1 mu^{2} cos ^{2} q + 1 mu^{2} sin ^{2} q
2 2
= 1 mu^{2} = Energy at the point of projection.
2
This is in accordance with the law of conservation of energy.
Problem 60. A projectile is projected with a kinetic energy K. Its range is R. It will have the minimum kinetic energy, after covering a horizontal distance equal to [UPSEAT 2002]
(a) 0.25 R (b) 0.5 R (c) 0.75 R (d) R
Solution : (b) Projectile possess minimum kinetic energy at the highest point of the trajectory i.e. at a horizontal distance R / 2.
Problem 61. A projectile is fired at 30^{o} with momentum p. Neglecting friction, the change in kinetic energy when it returns to the ground will be
(a) Zero (b) 30% (c) 60% (d) 100%
Solution : (a) According to law of conservation of energy, projectile acquire same kinetic energy when it comes at same level.
Problem 62. A particle is projected making angle 45^{o} with horizontal having kinetic energy K. The kinetic energy at highest point will be [CBSE PMT 2000, 01; AIEEE 2002]
 K
K (c) 2K (d) K
2
Solution : (b) Kinetic energy at the highest point K‘ = K cos ^{2} q = K cos ^{2} 45 ^{o} = K / 2
Problem 63. Two balls of same mass are projected one vertically upwards and the other at angle 60^{o} with the vertical. The ratio of their potential energy at the highest point is
(a) 3 : 2 (b) 2 : 1 (c) 4 : 1 (d) 4 : 3
Solution : (c) Potential energy at the highest point is given by PE = 1 mu^{2} sin ^{2} q
2
For first ball q = 90° \ (PE)_{1} = 1 mu ^{2}
2
For second ball q = (90 ^{o} – 60 ^{o} ) = 30° from the horizontal \ (PE)_{2} = 1 mu ^{2} sin ^{2} 30°
2
= 1 mu^{2}
8
\ (PE)I
(PE)II
= 4 : 1
Problem 64. In the above problem, the kinetic energy at the highest point for the second ball is K. What is the kinetic energy for the first ball
 4 K (b) 3 K (c) 2 K (d) Zero
Solution : (d) KE at the highest point KE = 1 mu^{2} cos ^{2} q
2
For first ball q = 90^{o} \ KE = 0
Problem 65. A ball is thrown at an angle q with the horizontal. Its initial kinetic energy is 100 J and it becomes 30 J at the highest point. The angle of projection is
(a) 45^{o} (b) 30^{o} (c) cos^{–1} (3/10) (d)
cos ^{1}(
3 / 10)
Solution : (d) KE at highest point
K‘ = K cos ^{2} q
2 2 3
1 æ 3 ö
30 = 100 cos q Þ
Horizontal Projectile.
cos
q = Þ q = cos
10
ç ÷


ç 10 ÷
A body be projected horizontally from a certain height ‘y’ vertically above the ground with initial velocity u. If friction is considered to be absent, then there is no other horizontal force which can affect the horizontal motion. The horizontal velocity therefore remains constant and so the object covers equal distance in horizontal direction in equal intervals of time.
 Trajectory of horizontal projectile : The horizontal displacement x is governed by the equation
x = ut Þ t = x
u
The vertical displacement y is governed by (since initial vertical velocity is zero)
…. (i)
y = 1 gt^{2} …. (ii)
2
By substituting the value of t in equation (ii) y = 1 g x 2
2 u 2
Problem 66. An aeroplane is flying at a constant horizontal velocity of 600 km/hr at an elevation of 6 km towards a point directly above the target on the earth’s surface. At an appropriate time, the pilot releases a ball so that it strikes the target at the earth. The ball will appear to be falling [MP PET 1993]
 On a parabolic path as seen by pilot in the plane
 Vertically along a straight path as seen by an observer on the ground near the target
 On a parabolic path as seen by an observer on the ground near the target
 On a zigzag path as seen by pilot in the plane
Solution : (c)
The path of the ball appears parabolic to a observer near the target because it is at rest. But to a Pilot the path appears straight line because the horizontal velocity of aeroplane and the ball are equal, so the relative horizontal displacement is zero.
Problem 67. The barrel of a gun and the target are at the same height. As soon as the gun is fired, the target is also released. In which of the following cases, the bullet will not strike the target
 Range of projectile is less than the initial distance between the gun and the target
 Range of projectile is more than the initial distance between the gun and the target
 Range of projectile is equal to the initial distance between the gun and target
 Bullet will always strike the target
Solution : (a) Condition for hitting of bullet with target initial distance between the gun and target £ Range of projectile.
Problem 68. A ball rolls off top of a staircase with a horizontal velocity u m/s. If the steps are h metre high and b mere wide, the ball will just hit the edge of nth step if n equals to
(a)
hu ^{2}
gb ^{2}
(b)
u ^{2} 8
gb ^{2}
(c)
2hu ^{2}
gb ^{2}
(d)
2u^{2} g
hb ^{2}
Solution : (c) By using equation of trajectory y = gx 2
2u ^{2}
for given condition
nh = g(nb)^{2}
2u ^{2}
\ n = 2hu ^{2}
gb ^{2}
 Displacement of Projectile (r ) : After time t, horizontal displacement x = ut and vertical displacement y = 1 gt ^{2} .
2
So, the position vector r = ut ˆi – 1 gt ^{2} ˆj
Therefore
r
r = ut
2
and
a = tan
_{1} æ gt ö
ç 2u ÷
è ø
1 æ gy ö æ ö






a = tan ç ÷
è ø
ç as t = ÷
è ø
 Instantaneous velocity : Throughout the motion, the horizontal component of the velocity is v_{x} = u.
The vertical component of velocity increases with time and is given by
v_{y} = 0 + g t = g t (From v = u + g t)
So,
i.e.
r
v = v
v =
_{x}ˆi – vy ˆj
r

= v = ui – g t j
= u
Again
i.e.
v = ui – ˆj


v =


v_{y} 1 æ v_{y} ö
1 æ
2gy ö
_{1} æ gt ö
Direction of instantaneous velocity : tan f = Þ f = tan
ç ÷ = tan ç ÷
or f = tan ç ÷
vx
Where f is the angle of instantaneous velocity from the horizontal.
è vx ø ç ÷
è u ø
Problem 69. A body is projected horizontally from the top of a tower with initial velocity 18 ms^{–1}. It hits the ground at angle 45^{o}. What is the vertical component of velocity when it strikes the ground
(a) 9 ms^{–1} (b) 9 ms^{–1} (c) 18 ms^{–1} (d) 18
Solution : (c) When the body strikes the ground
tan 45 ^{o} = vy = vy = 1
ms–1
vx 18
vy = 18 m / s.
Problem 70. A man standing on the roof of a house of height h throws one particle vertically downwards and another particle horizontally with the same velocity u. The ratio of their velocities when they reach the earth’s surface will be
(a) : u (b) 1 : 2 (c) 1 : 1 (d) :
Solution : (c) For first particle : v ^{2} = u ^{2} + 2gh Þ v =
For second particle : v = = =
So the ratio of velocities will be 1 : 1.
Problem 71. A staircase contains three steps each 10 cm high and 20 cm wide. What should be the minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane
(a) 0.5 m/s (b) 1 m/s (c) 2 m/s (d) 4 m/s
Solution : (c) Formula for this condition is given by
n = 2hu ^{2}
gb ^{2}
where h = height of each step, b = width of step, u = horizontal velocity of projection, n = number of step.
Þ 3 = 2 ´ 10 ´ u 2 Þ u ^{2} = 200 cm / sec = 2 m / sec
10 ´ 20^{2}
(4) Time of flight : If a body is projected horizontally from a height h with velocity u and time taken by the body to reach the ground is T, then
h = 0 + 1 gT ^{2}
2
(for vertical motion)
T =
Problem 72. Two bullets are fired simultaneously, horizontally and with different speeds from the same place. Which bullet will hit the ground first [Orissa JEE 2003]
 The faster one (b) Depends on their mass
(c) The slower one (d) Both will reach simultaneously
Solution : (d)
Problem 73. An aeroplane is flying at a height of 1960 m in horizontal direction with a velocity of 360 km/hr. When it is vertically above the point. A on the ground, it drops a bomb. The bomb strikes a point B on the ground, then the time taken by the bomb to reach the ground is
Solution : (b)
(a) 20
t =
sec (b) 20 sec (c)
= = 20 sec
10 sec (d) 10 sec
 Horizontal range : Let R is the horizontal distance travelled by the body
R = uT + 1
2
0 T ^{2}
(for horizontal motion)
R = u
Problem 74. A bomb is dropped on an enemy post by an aeroplane flying with a horizontal velocity of 60 km/hr and at a height of 490 m. How far the aeroplane must be from the enemy post at time of dropping the bomb, so that it may directly hit the target. (g = 9.8 m/s^{2})
(a)
100 m
3
(b)
500 m
3
(c)
200 m
3
(d)
400 m
3
Solution : (b)
S = u ´ t = u ´
= 60 ´ 5 ´
18
2 ´ 490
9.8
= 500 m
3
Problem 75. A body is thrown horizontally with velocity from the top of a tower of height h. It strikes the level ground through the foot of tower at a distance x from the tower. The value of x is
 h (b)
h (c) 2 h (d) 2h
2 3
Solution : (c) x = u ´ = ´
\ x = 2h
Problem 76. An aeroplane moving horizontally with a speed of 720 km/h drops a food packet, while flying at a height of 396.9 m. The time taken by a food packet to reach the ground and its horizontal range is (Take g = 9.8 m/sec^{2}) [AFMC 2001]
 3 sec and 2000 m (b) 5 sec and 500 m (c) 8 sec and 1500 m (d) 9 sec and 1800 m
Solution : (d) Time of descent t = =
Þ t = 9 sec
and horizontal distance S = u ´ t
S = æ 720 ´ 5 ö ´ 9
= 1800 m


ç 18 ÷
(6) If projectiles A and B are projected horizontally with different initial velocity from same height and third particle C is dropped from same point then
 All three particles will take equal time to reach the
(ii) Their net velocity would be different but all three particle possess same vertical component of velocity.
 The trajectory of projectiles A and B will be straight line r.t. particle C.
(7)
If various particles thrown with same initial velocity but indifferent direction then
 They strike the ground with same speed at different times irrespective of their initial direction of
(ii) Time would be least for particle E which was thrown vertically downward.
 Time would be maximum for particle A which was thrown vertically
Projectile Motion on an Inclined Plane.
Let a particle be projected up with a speed u from an inclined plane which makes an angle a with the horizontal velocity of projection makes an angle q with the inclined plane.
We have taken reference xaxis in the direction of plane.
Hence the component of initial velocity parallel and perpendicular to the plane are equal to u cos q and u sin q
respectively i.e. u = u cosq
and
u_{^} = u sinq .
The component of g along the plane is
g sina
and perpendicular to
the plane is
a^ = g cosa .
g cosa
as shown in the figure i.e.
a = –g sin a
and
Therefore the particle decelerates at a rate of from O to P.
g sina
as it moves
 Time of flight : We know for oblique projectile motion T = 2u sinq
g
or we can say
T = 2u^
a^
\ Time of flight on an inclined plane T = 2u sinq
g cosa
 Maximum height : We know for oblique projectile motion
u2
H = u^{2} sin ^{2} q
2g
or we can say
H = ^
2a^
\ Maximum height on an inclined plane
H = u^{2} sin ^{2} q
2g cosa
 Horizontal range : For one dimensional motion
s = ut + 1 at ^{2}
2
Horizontal range on an inclined plane
R = u T + 1 a T ^{2}
R = u cosq T – 1 g sina T ^{2}
2
 2 
æ 2u sinq ö 1
æ 2u sinq ö ^{2}
R = u cosq ç ÷ – g sina ç ÷
g cosa 2 g cosa
è ø è ø
By solving R = 2u2
g
sinq cos(q + a) cos ^{2} a
(i) Maximum range occurs when q = p – a
4 2
(ii) The maximum range along the inclined plane when the projectile is thrown upwards is given by
Rmax
= u 2
g (1 + sin a)
(iii) The maximum range along the inclined plane when the projectile is thrown downwards is given by
Rmax
= u 2
g (1 – sin a)
Problem 77. For a given velocity of projection from a point on the inclined plane, the maximum range down the plane is three times the maximum range up the incline. Then, the angle of inclination of the inclined plane is
(a) 30^{o} (b) 45^{o} (c) 60^{o} (d) 90^{o}
Solution : (a) Maximum range up the inclined plane (R
max
)up
= u 2
g(1 + sin a)
Maximum range down the inclined plane (R
max
)down
= u 2
g(1 – sin a)
and according to problem : By solving a = 30^{o}
u 2
g(1 – sin a)
= 3 ´
u 2
g(1 + sin a)
Problem 78. A shell is fired from a gun from the bottom of a hill along its slope. The slope of the hill is a = 30^{o}, and the angle of the barrel to the horizontal b = 60^{o}. The initial velocity v of the shell is 21 m/sec. Then distance of point from the gun at which shell will fall
(a) 10 m (b) 20 m (c) 30 m (d) 40 m Solution : (c) Here u = 21 m/sec, a = 30^{o}, q = b – a = 60^{o} – 30^{o} = 30^{o}
Maximum range R = 2u 2
sinq cos(q + a) = 2´ (21)^{2} ´ sin 30 ^{o} cos 60 ^{o}
= 30 m
g cos ^{2} a 9.8 ´ cos ^{2} 30 ^{o}
Problem 79. The maximum range of rifle bullet on the horizontal ground is 6 km its maximum range on an inclined of 30^{o} will be
 1 km (b) 2 km (c) 4 km (d) 6 km
Solution : (c) Maximum range on horizontal plane R = u 2 = 6km (given)
g
Maximum range on a inclined plane R
max
= u 2
g (1 + sina)
Putting a = 30^{o}
Rmax =
2 æ 2 ö







ç ÷
o
2 ´ 6 = 4 km.

g (1 + sin 30 ) ç ÷