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� projectile

An object thrown into space with certain velocity, fired from a gun or dropped from a moving plane is called projectile.

A projectile moves with a constant horizontal velocity and at the same time falls freely under the action of gravity. The path of projectile is called trajectory.

� Trajectory of a Projectile

Let a particle of mass ‘m’ is projected from a point ‘O’ with initial velocity ‘v0’ making an angle ‘α’ with horizontal. Take ‘O’ as origin and horizontal and vertical lines through ‘O’ as x-axis and y-axis respectively.

y

P(x, y)

v0

r�

α � mg j�

0

x-axis

Suppose that after time ‘t’ the particle is at point P(x, y) whose position vector is r� . i.e.

r� = xi� + yj� ⇒

dr�dt

=dx

dti� +

dy

dtj�

⇒ v�� = dx

dti� +

dy

dtj�

CHAPTER

7 PROJECTILE MOTION


2

⇒ dv��dt

=d

2x

dt2i� +

d2y

dt2j�

⇒ a� = d2x

dt2i� +

d2y

dt2j� ________(i)

The gravitational force F�� acting on the particle at point P(x, y) is

F�� = � mg j� ________(ii)

By Newton’s 2nd law of motion

F�� = ma� = m �d

2x

dt2i� +

d2y

dt2j�� ________(iii)

From (ii) & (iii), we get

m �d2x

dt2i� +

d2y

dt2j�� mg j�

⇒ d2x

dt2i� +

d2y

dt2j� 0.i� � g j�

⇒ d2x

dt2 0 and d

2y

dt2 � g

On integrating with respect to “t”, we get dx

dt= A and

dy

dt= � gt + B _________(iv)

Where A & B are constant of integration. To determine the value of these constant we apply the initial conditions. Initially at t = 0

dx

dt= v0cosα and

dy

dt= v0sinα

Using these values in (iv), we get

A = v0cosα and B = v0sinα

Using values of A & B in (iv), we get

dx

dt= v0cosαααα and dy

dt= v0sinαααα � gt ________(v)

Eq(v) gives the horizontal and vertical components of velocity at any time t. On integrating (v), with respect to ‘t’, we get

x = v0cosα�t + C and y = v0sinα�t � 1

2gt2 + D ________(vi)

Where C & D are constant of integration. To determine the value of these constant we apply

the initial conditions.

Initially at t = 0, x = 0 and y = 0


3

⇒ C = D = 0

Using value of C & D in (vi), we get

x = v0cosαααα��t ________(vii) and y = v0sinαααα��t � 1

2gt2 ________(viii)

Equations (vii) and (viii) are parametric equations of trajectory. Now we find Cartesian equation of trajectory. From (vii)

t = x

v0cosα

Putting value of t in (viii), we get

y = v0sinα� x

v0cosα� 1

2g � x

v0cosα 2

⇒ y = xtanαααα � gx2

2v02sec2αααα

Which is Cartesian equation of trajectory of a projectile.

� Vertex, Latus Rectum & Maximum Height of a Projectile

We know that the Cartesian equation of the trajectory of a projectile is:

y = xtanα � gx2

2v02

sec2α

⇒ gx2

2v02

sec2α = xtanα � y

⇒ x2 = � 2v02

gsec2α� xtanα � y � 2v0

2

gsec2α�

⇒ x2 = xv0

2sinαcosαg

� 2yv0

2cos2α

g

⇒ x2 � xv0

2sinαcosαg

� 2yv0

2cos2α

g

Adding �v02sinαcosα

g�2

on both sides we get

x2 � xv0

2sinαcosαg

+ �v02sinαcosα

g�2 �v0

2sinαcosα

g�2 � 2yv

02cos2α

g

⇒ �x � v02sinαcosα

g�2

= � 2v02cos2α

g�y � v0

2sin2α

2g�

Comparing with (x – h)2 = 4a(y – k), we get


4

h v02sinαcosα

g, 4a � 2v0

2cos2α

g, k v0

2sin2α

2g

Thus, Vertex = h, k� �v02sinαcosα

g, v0

2sin2α

2g �

Latus Rectum = ||||4a|||| 2v02cos2α

g

Height (H) = k v02sin2α

2g

� Focus

X-coordinate of focus = x-coordinate of vertex

v02sinαcosα

g

v022sinαcosα

2g

v02

2gsin2α

Y-coordinate of focus = H � 1

4(Latus Rectum)

v02sin2α

2g� 1

4�2v0

2cos2α

g�

v02sin2α

2g� v0

2cos2α

2g

� v02

2gcos2α � cos2α� � v0

2

2gcos2α

Thus, Focus = �v02

2gsin2α, � v0

2

2gcos2α�

� Equation of Directrix

Height of directrix above the x-axis is:

y = H + 1

4(Latus Rectum)

v02sin2α

2g+

1

4�2v0

2cos2α

g�


5

v02sin2α

2g+

v02cos2α

2g

v02

2gsin2α + cos2α�

v02

2g

� Time of Flight

The time taken by the projectile in reaching the final point is called the time of flight of the projectile. We know that parametric equation of trajectory of projectile are:

x = v0cosα�t and y = v0sinα�t � 1

2gt2

To find the time of flight put y = 0

v0sinα�t � 1

2gt2 = 0

⇒ �v0sinα � 1

2gt� t = 0

⇒ v0sinα � 1

2gt = 0 � t ≠ 0

⇒ t = 2v0sinα

g

Thus, T.F = 2v0sinα

g

� Range of a Projectile

The range or horizontal range of the projectile is the horizontal distance covered by the projectile during time of flight.

Range (R) = (Horizontal Velocity)(Time of Flight)

v0cosα� �2v0sinαg

� v02

gsin2α

R will be maximum when sin2α is maximum.

i.e. sin2α = 1

⇒ 2α = sin – 1(1)

⇒ 2α = 900

⇒ α = 450


6

Which shows that if projectile is projected with an angle of 450 then it covers the maximum horizontal distance.

Thus Rmax= v02

g

� Question 1

Determine the maximum possible range for a projectile fired from a cannon having muzzle velocity v0 and prove that the height reached in this case is

v02

4g

Solution

� Solution We know that

Rage = (Horizontal Velocity)(Time of Fight)

v0cosα� �2v0sinα

g� v0

2

gsin2α


i.e. sin2α = 1 ⇒ 2α = sin – 1(1)

⇒ 2α = 900

⇒ α = 450

Which shows that if projectile is projected with an angle of 450 then it covers the maximum horizontal distance.

Thus Rmax= v02

g

As Height reached v02sin2α

2g

Put α = 450

Height reached v02sin245

2g v0

2 � 1√2�2

2g v0

2

4g

� Question 2

What is the maximum range of possible for a projectile fired from a cannon having muzzle velocity 1mile/sec. What is the height reached in this case.?

Solution


7

We know that



g� v0

2

gsin2α


i.e. sin2α = 1

⇒ 2α = sin – 1(1)

⇒ 2α = 900

⇒ α = 450

So Rmax= v02

g

Given that v0 = 1 mile/sec = 1760 yard/sec = 1760 × 3 ft/sec = 5280ft/sec

Thus Rmax=5280

2

32

871200 feet 871200

5280 mile 165 mile


2g

Put α = 450

Height reached v02sin245

2g

v02 � 1√2

�2

2g v0

2

4g

52802

4(32)

217800 feet

217800

5280 mile 41.25 mile

� Question 3

A cannon has its maximum range R. Prove that

(a) the height reached is R4�

(b) the time of flight is �2Rg�

Solution


8

We know that



g� v0

2

gsin2α


i.e. sin2α = 1 ⇒ 2α = sin – 1(1) ⇒ 2α = 900 ⇒ α = 450

So Rmax= v02

g

Given that Rmax = R

⇒ R = v02

g

⇒ v02 = Rg


2g

Rg sin245

2g R � 1√2

�2

2 R

4

We know that

Time of Flight 2v0sinα

g

=2�Rgsin45

g=

2�Rg

g√2=�2R

g

� Question 4

A projectile having horizontal range T, reaches a maximum height H. Prove that it must have been launched with

(a) an initial speed equal to

�g�R2+16H2�8H

(b) at an angle with horizontal given by

sin�1 � 4H�R2+16H

2�

Solution

We know that

R = Rage = (Horizontal Velocity)(Time of Fight)


9


g� v0

2

gsin2α

and H = Height reached v02sin2α

2g

Now

R2+16H2= �v02

gsin2α�2

+ 16 �v02sin2α

2g�2

= v0

4

g24sin2αcos2α + 4

v04sin4α

g2

= 4v0

4

g2sin2α�cos2α + sin

2α�

= 4v0

4sin2α

g2 ________(i)

⇒ R2+16H2

8H=

4v04sin2α

g2

g

4v02sin2α

= v0

2

g

⇒ v02 = g�R2+16H2�

8H

⇒ v0 = �g�R2+16H2�8H

From (i)

�R2+16H2 = 2v02sinα

g

⇒ 4H�R2+16H2

= 2v02sin2α

g

g

2v02sinα

⇒ sinα 4H�R2+16H2⇒ α sin�1 � 4H�R2+16H2

�

� Question 5

Find the range of a rifle bullet when α is the elevation of projection and v0 the speed. Show that, if the rifle is fired with the same elevation and the speed from a car travelling with speed V towards the target, and the range will be increased by

2v0Vsinαααα

g

Solution


10

We know that


⇒ R v0cosα� �2v0sinα

g� v0

2

gsin2α

When shell is fired from a car moving with velocity V towards the target then the horizontal velocity increased by V.

i.e. Horizontal velocity = v0cosα + V

Let R� be new range. Then R� = (New Horizontal Velocity)(Time of Fight)

v0cosα + V� �2v0sinα

g�

v02

gsin2α + 2v0Vsinα

g

Now

Increased in Range = R� � R

v02

gsin2α + 2v0Vsinα

g� v0

2

gVsin2α

2v0Vsinα

g

� Question 6

The range of a rifle bullet is 1200yards when α is the elevation of projection. Show that, if the rifle is fired with the same elevation and the speed from a car travelling at 10 miles per hour towards the target the range will be increased by 220√tanαααα feet. Solution

Given that

R = 1200yards = 1200 × 3 = 3600ft.

and V = 10 mile /h

= 10 ×1760 ×3

3600

= 44

3ft/sec

We know that



11

⇒ R v0cosα� �2v0sinα

g� v0

2

gsin2α

⇒ v02 Rg

sin2α

⇒ v0 � Rg

sin2α � Rg

2sinαcosα

When shell is fired from a car moving with velocity V towards the target then the horizontal velocity increased by V. i.e. Horizontal velocity = v0cosα + V

Let R� be new range. Then

R� = (New Horizontal Velocity)(Time of Fight)

v0cosα + V� �2v0sinα

g�

v02

gsin2α + 2v0Vsinα

g

Now

Increased in Range = R� � R

v02

gsin2α + 2v0Vsinα

g� v0

2

gVsin2α

2v0Vsinα

g

2Vsinα

g� Rg

2sinαcosα

V�2Rg √tanα

44

3�2 × 3600

32 √tanα

44

3 60

4√tanα

220√tanα


12

� Question 7

A battleship is steaming ahead with sped V and a gun is mounted on the battleship so as the point straight backwards, and is set at an angle of elevation α. If v0 is the speed of projection (relative to the gun), show that the range is

2v0

gsinαααα v0cosαααα � V�

Also prove that the angle of elevation for maximum range is

cos� 1 !"V +�V2+ 8v0

2

4v0 #$

Solution

When shell is fired from a battleship moving ahead with velocity V towards the target which is behind the battleship then the horizontal velocity decreased by V.

i.e. Horizontal velocity = v0cosα � V

Let R be the range. Then

R = (Horizontal Velocity)(Time of Fight)

v0cosα � V� �2v0sinα

g�

2v0sinα

gv0cosα � V�

Differentiating w.r.t “α”, we get dR

dα 2v0

g%cosαv0cosα � V� & sinα�v0sinα �'

2v0

g(v0cos2α � Vcosα � v0sin

2α)

2v0

g(v0�cos2α�sin

2α� � Vcosα)

2v0

g*v0 �cos2α � 1 �cos2α� � Vcosα+

2v0

g%v02cos2α � 1� � Vcosα'

2v0

g%2v0cos2α � Vcosα � v0'

Differentiating again w.r.t “α”, we get

d2R

dα2 2v0

g%�4v0cosαsinα + Vsinα'


13

2v0

gsinα%V � 4v0cosα'

Putting dR

dα= 0, we get

2v0

g%2v0cos2α � Vcosα � v0' 0

⇒ 2v0cos2α � Vcosα � v0 0

⇒ cosα = V ± �V2 + 8v02

4v0

At cosα = V + �V2 + 8v02

4v0

d2R

dα2 2v0

gsinα -V � 4v0

V + �V2 + 8v02

4v0

. 2v0

gsinα *��V2 + 8v0

2+ So

d2R

dα2 < 0 at cosα =

V + �V2 + 8v02

4v0

Which shows that R is maximum at cosα = V + �V2 + 8v0

2

4v0

. Thus the angle of elevation for maximum range is given by

cos� 1 0V + �V2 + 8v02

4v0

1

� Question 8

A shell bursts on contact with the ground and pieces from it fly in all directions with all speeds up to 80feet per seconds. Prove that a man 100 feet away is in danger for 5 √2

�

seconds.

Solution

We know that



g�

v02

gsin2α


14

Given that

R = 100ft, v0 = 80ft/sec and g = 32ft/sec2

So 100 =80

2

32sin2α

⇒ 100 =6400

32sin2α

⇒ sin2α =1

2 ⇒ 2α = sin�1 �1

2� ⇒ 2α = 30, 150 ⇒ α = 15, 75

For the range of 100ft. there are two angles of projection. Let T1 and T2 be the times of the flights respectively. Then

T1=2v0sin15

g and T2=

2v0sin75

g

Let T be the maximum time of danger for the man. Then T = T2 – T1

= 2v0sin75

g� 2v0sin15

g

= 2v0

gsin75 � sin15�

= 2v0

g2cos �75 + 15

2� sin �75 � 15

2�

= 2v0

g2cos45sin30

= 4(80)

32� 1√2

� �1

2� =

5√2sec

� Question 9

A number of particles are projected from the same point at the same instant in various directions with speed v0. Prove that at any subsequent time t, they will be on a sphere of radius v0t and determine the motion of the centre of the sphere.

Solution

Let a particle moving with velocity v0 makes an angle α. Let after time ‘t’ a particle is at a point P(x, y, z). Then

x = (v0cosα)t

y = (v0sinα) t – 1

2 gt

2 ⇒ y +1

2 gt

2 = (v0sinα) t

T2

750

T1

150 100ft.


15

z = 0

Squaring and adding we get

x2 + � y +

1

2 gt

2�2

+ z2= (v0cosα)t�2 +(v0sinα)t�2

= v02t2�cos2α + sin

2α� = v0t�2

Which is a sphere of a radius v0t centered at � 0, � 1

2 gt

2, 0 . Since the centre lies on the

vertical axis and as t increases centre descends under gravity along vertical axes.

� Question 10

Prove that the speed required to project a particle from a height h to fall a horizontal distance a from the point of projection is at least

�g ��a2+ h2 � h

Solution v0 α

x

O a h P(a, �h)

Let O be the point of the projection from where the projectile is projected. Let v0 be the velocity making angle α with horizontal. Let h be the height of the point of the projection O and projectile fall a distance a from O. Let it falls at a point P, therefore the coordinates of P are (a, � h).


y = xtanα � gx2

2v02

sec2α

Since P(a, � h) lies on it, therefore

� h = atanα � ga2

2v02

sec2α

⇒ � 2v02h = 2av0

2tanα � ga2�1 & tan2α�


16

⇒ ga2tan2α � 2v02xtanα + ga2 � 2v0

2h = 0

Since it is coodratic in tanα and tanα is real therefore discriminate is greater than zero. i.e.

b2 – 4ac ≥ 0

⇒ 2av02�3 � 4ga2�ga2 � 2v0

2h� ≥ 0

⇒ 2av02�3 ≥ 4ga2�ga2 � 2v0

2h�

⇒ 4a2v0

4 ≥ 4g2a4 � 8gha2v0

2 ⇒ v0

4 ≥ g2a2 � 2ghv02

⇒ v04 + 2ghv0

2 ≥ g2a2 ⇒ v04 + 2ghv0

2+ gh�2 ≥ g2a2 + gh�2

⇒ v02 + gh�3 ≥ g2�a2 & h2�

⇒ v02 + gh ≥ �g2�a2 & h2�

⇒ v02 ≥ g�a2 & h2 � gh

⇒ v0 ≥ �g ��a2 & h2 � h

Hence the least velocity of projection is

v0 = �g ��a2 & h2 � h

� Question 11

A projectile is launched at an angle α from a cliff of height H above the see level. if it falls into the sea at a distance D from the base of the cliff, prove that the maximum height above sea level is

H + D2 tan2αααα

4H + Dtanαααα� Solution v0 h α

x

O D H P(D, �H)


17

Let O be the point of the projection from where the projectile is projected. Let v0 be the velocity making angle α with horizontal. Let H be the height of the point of the projection O and projectile fall a distance D from O. Let it falls at a point P, therefore the coordinates of P are (D, � H). Let h be the height above the x-axis then

h = v0

2sin2α

2g ________(i)


y = xtanα � gx2

2v02

sec2α

Since P(D, � H) lies on it, therefore

� H = Dtanα � gD2

2v02

sec2α

⇒ Dtanα + H = gD2

2v02

sec2α

⇒ v02 =

gD2

2H + Dtanα�cos2α

Using value of v02 in (i), we get

h = � gD2

2H + Dtanα�cos2α� sin

2α

2g

= � D2tan2α

4H + Dtanα��

Height above sea level = H + h

H + D2tan2α

4H + Dtanα�

� Question 12

A ball is dropped from the top of a tower of height h. At the same moment, another ball is thrown from a point of the ground at a distance k from the foot of tower so as to strike the first ball at the depth d. Show that the initial speed and the direction of projection of the speed ball are respectively

�g�h2 � k2�2d

and tan� 1 �hk

�

Solution


18

Let the first ball is dropped from the height h and it strikes the second ball at the depth d at a point P whose coordinates are (k, h – d).

We know that

x = ut + 1

2gt2

For 1st ball, x = d and u = 0

So d = 1

2gt2 ________(i)

For 2nd ball parametric equations are

x = (v0cosα)t

and y = (v0sinα)t � 1

2gt2

Since P(k, h – d) lies on it therefore

k = (v0cosα)t ______(ii)

and h � d = (v0sinα)t � 1

2gt2 ______(iii)

Adding (i) and (iii), we get

h = (v0sinα)t ______(iv)

Squaring (ii) and (iv) then adding, we get

h2+ k

2 = (v0sinα)t�2+ (v0cosα)t�2

= v02t

2sin2α + cos2α� = v02t2

= v02

2d

g By(i)

⇒ v02 g�h

2+ k

2�2d

⇒ v0 �g�h2+ k

2�2d

I

d

P h – d h v0 α

x II O k


19

From (ii) & (iv), we have

tanα =h

k ⇒ α = tan�1 �h

k�

� Question 13

From a gun placed on a horizontal plane, which can fire a shell with speed �2gH, it is required to thro a shell over a wall of height h, and the elevation of the gun cannot exceed α < 450. Show that this will be possible only when h < Hsin2α, and that, if this condition be satisfied, the gun must be fired from within a strip of the plane whose breadth is

4cosαααα �HHsin2αααα � h� Solution y

C

D

v0

h

h

α

x O� O A B

Let AC be a wall of height h and particle be projected at O with speed v0 making an angle α.

Then v0 = �2gH (given)

Now for a shell to cross the wall, the height of the wall is less than the height of vertex.

i.e. h < v0

2sin2α

2g

⇒ h < 2gHsin2α

2g � v0 = �2gH

⇒ h < Hsin2α

Which is required.


y = xtanα � gx2

2v02

sec2α

Putting y = h and v0 = �2gH, we get


20

h = xtanα � x2

4Hsec2α

⇒ 4hH = 4xHtanα � x2sec2α

⇒ x2sec2α � 4xHtanα + 4hH 0

⇒ x2 � 4xHtanαcos2α + 4hHcos2α 0

⇒ x2 � 4xHsinαcosα + 4hHcos2α 0

⇒ x =

4Hsinαcosα ±�16H2sin2α cos2α � 41�(4hHcos2α)

2

= 4Hsinαcosα ±�16H2sin2α cos2α � 16hHcos2α

2

= 4Hsinαcosα ± 4cosα�H2sin2α � hH

2

= 2Hsinαcosα ± 2cosα�H2sin2α � hH

Thus

OB = 2Hsinαcosα + 2cosα�H2sin2α � hH

OA =2Hsinαcosα � 2cosα�H2sin2α � hH

Hence the breadth of the strip is:

OO� = AB = OB – OA = 4cosα�H2sin2α � hH


21

� Question 14

A shell fired with speed V at an elevation θ, hits an airship at height H,, which is moving horizontally away from the gun with speed v0. Show that, if 2Vcosθθθθ � v0��V2

sin2θθθθ � 2gH� = v0Vsinθθθθ

The shell might also have hit the air ship if the latter had remained stationary in the position it occupied when the gun was actually fired.

Solution y A B v

H

H

α

x O Let A be the position of airship when shot was fired and it hit plane at B. If t is the time taken

by hell to reach height H.

We know that

y = (v0sinα)t � 1

2gt2

Putting v0 = V, θ = α and y = H, we get

H = (Vsinθ)t � 1

2gt2

⇒ 2H = (Vsinθ)t � gt2

⇒ gt2 � (Vsinθ)t & 2H = 0 ⇒ t =

2Vsinθ ±�4V2sin2θ � 8Hg2g

=

Vsinθ ±�V2sin2θ � 2Hgg


22

Let t1 and t2 be the time taken by the shell to reach the point A and B respectively. Then

t1 =

Vsinθ � �V2sin2θ � 2Hg

g and t2 =

Vsinθ +�V2sin2θ � 2Hg

g

Let T be the time taken by the shell from A to B. Then

T = t2 – t1

Vsinθ +�V2sin2θ � 2Hg

g� Vsinθ � �V2sin2θ � 2Hg

g

2�V2sin2θ � 2Hg

g

Now |AB| = Distance covered by shell

= (Horizontal Velocity)(Time)

Vcosθ� 2�V2sin2θ � 2Hg

g

2Vcosθ�V2sin2θ � 2Hg

g ______(i)

Now |AB| = Distance covered by airship

= (Velocity)(Time) = v0t2

v0 �Vsinθ +�V2sin2θ � 2Hg�g

______(ii)

From (i) and (ii), we get

2Vcosθ�V2sin2θ � 2Hg

g v0 �Vsinθ +�V2sin2θ � 2Hg�

g

⇒ 2Vcosθ�V2sin2θ � 2Hg v0Vsinθ + v0�V2sin2θ � 2Hg

⇒ v0Vsinθ = 2Vcosθ�V2sin2θ � 2Hg � v0�V2sin2θ � 2Hg

⇒ v0Vsinθ = 2Vcosθ � v0� �V2sin2θ � 2Hg Which is required.


23

� Question 15

An aeroplane is flying with constant speed v0 and at constant height h. Show that if, a gun is fired point blank at the aeroplane after it has passed directly over the gun when its angle of elevation as seen from the gun is α, the shell will hit the aero plane provided that

2Vcosαααα � v0�� v0tan2αααα = gh

Where V is the initial speed of the shot, the path being assumed parabolic.

Solution

A

B

h α O C D

Let A be the position of plane when shot was fired and it hit plane at B. Let v0 be the speed of plane. From fig.

AB = v0t ( � S = vt)

Horizontal coordinate of B = OD

= OC + CD

= OC + AB � AB = CD

= OC + v0t � AB = v0t _____(i)

In ∆AOC

AC

OC= tanα

⇒ h

OC= tanα ⇒ OC h

tanα ⇒ OC hcotα

Using value of OC in (i), we get

Horizontal coordinate of B = hcotα + v0t

Thus coordinates of B are (hcotα + v0t, h)

The parametric equations are:

x = (v0cosα)t


2gt2


24

Here v0 = V therefore

x = (Vcosα)t ______(ii)

y = (Vsinα)t � 1

2gt2 ______(iii)

Since B lies on the trajectory therefore x = hcotα + v0t and y = h

Using x = hcotα + v0t in (ii), we get

hcotα + v0t = (Vcosα)t

⇒ hcotα = (Vcosα)t � v0t

⇒ hcotα = (Vcosα � v0)t

⇒ t = hcotα

Vcosα � v0

Using y = h in (iii), we get

h = (Vsinα)t � 1

2gt2

= (Vsinα) � hcotα

Vcosα � v0

� 1

2g � hcotα

Vcosα � v0

2

= � Vhcosα

Vcosα � v0

� � 1

2g � hcotα

Vcosα � v0

�2

⇒ 1 = � Vcosα

Vcosα � v0

� � h

2g � cotα

Vcosα � v0

�2

⇒ 2Vcosα � v0�2 = 2Vcosα�Vcosα � v0� � hgcot2α

⇒ ghcot2α = 2Vcosα�Vcosα � v0� � 2Vcosα � v0�2

⇒ ghcot2α = 2Vcosα � v0�Vcosα � Vcosα + v0�

⇒ gh = 2v0Vcosα � v0�tan2α


25

� Parabola of Safety

A parabola which touches the every trajectory of a projectile which is formed inward for different value of angle of projection with same initial velocity v0 is called parabola of safety.

y

Parabola of safety

x

O

Equation of trajectory of a parabola is:

y = xtanα � gx2

2v02

sec2α

⇒ y = xtanα � gx2

2v02

1 + tan2α�

⇒ y = xtanα � gx2

2v02

� gx2

2v02

tan2α

⇒ gx2

2v02

tan2α � xtanα + � gx2

2v02

+ y� = 0

This equation is quadratic in tanα. For envelope put discriminate of equation equal to zero. i.e.

b2 – 4ac = 0

⇒ x2 � 4 � gx2

2v02� � gx2

2v02

+ y� = 0

⇒ 1 � 2 � g

v02� �gx2+ 2v0

2y

2v02

� = 0

⇒ g �gx2+ 2v02y

v04

� = 1

⇒ gx2+ 2v02y = v0

4

g

⇒ gx2 = � 2v02y + v0

4

g


26

⇒ x2 = � 2v02y

g +

2v04

2g2

⇒ x2 = � 2v02

g�y �

v02

2g�

Which is the equation of parabola of safety.

� Question 16

A particle is projected at time t = 0 in a fixed vertical plane from a given point O with speed �2ga of which the vertical component is V. Show that at time t = 2a/V the particle is on a

fixed parabola (parabola of safety), that its path touches the parabola, and that its direction of motion is then perpendicular to its direction of projection.

Solution

Given that v0 = �2ga and V = v0sinα

We know that the equation of parabola of safety is:

x2 = � 2v02

g�y �

v02

2g�

⇒ x2 = � 2��2ga�2

g7y �

��2ga�2

2g8

= � 4ay � a� _______(i)

Which is the equation of the parabola of safety.

Let P(x, y) be a point on trajectory then x = (v0cosα)t ______(ii)


2gt2 ______(iii)

At t = 2a

V ii� becomes

x = (v0cosα)2a

V

= (v0cosα)2a

v0sinα

⇒ x = 2acotα ______(iv)

At t = 2a

V iii� becomes


27

y = (v0sinα)2a

V � 1

2g �2a

V�2

= (v0sinα)2a

v0sinα � 1

2g � 2a

v0sinα�2

= 2a � 1

2g � 2a�2ga sinα

�2

= 2a � a cosec2α ______(v)

Thus the coordinates of P are (2acotα, 2a � a cosec2α)

Putting (iv) and (v) in (i), we get 2acotα�2 = � 4a2a � a cosec2α � a�

⇒ 4a2cot2α = � 4aa � acosec2α�

⇒ a2cot2α = a2cosec2α � 1�

⇒ a2cot2α = a2cot2α

⇒ L.H.S = R.H.S

Thus P(x, y) lies on the parabola of safety. So at t = 2a/V this trajectory touches the parabola of safety.

Differentiate (ii) & (iii) w.r.t “t”, we get

dx

dt = v0cosα and dy

dt = v0sinα � gt

Now dy

dx=

dy

dt: dx

dt

= v0sinα � gt

v0cosα

At t = 2a

V

dy

dx= v0sinα � g 2a

V

v0cosα

= Vv0sinα � 2ag

Vv0cosα

= v0sinα�v0sinα � 2agv0sinα�v0cosα � V = v0sinα

= ��2gasinα��2gasinα � 2ag��2gasinα��2gacosα � v0 = �2ga


28

= 2agsin2α � 2ag

2agsinαcosα

= � 1 � sin2α

sinαcosα

= � cos2α

sinαcosα

= � cotα

This is the slope of the tangent at t = 2a/V

Since dy

dx= tanθ

Where θ is inclination of the tangent at P(x, y). Then

tanθ = � cotα tan(900+ α)

⇒ θ = 900+ α

Thus at t = 2a/V, its direction of motion is perpendicular to the direction of projection.

� Range of a Projectile on Inclined Plane

Let a plane be inclined at an angle β to the horizontal. Let a particle is projected from point O

with velocity v0 by making an angle α to the horizontal with β < α. Let the projectile meet the inclined plane at a point P(x, y). Then OP = r is called the range of projectile on inclined plane. Then x = rcosβ and y = rsinβ

So P(x, y) = (rcosβ, rsinβ)

y

P(x, y)

v0 r

α rsinβ

β

x O rcosβ

Equation of the trajectory is:

y = xtanα � gx2

2v02

sec2α


29

Since P lies on it therefore

rsinβ = rcosβtanα � gr2cos2β

2v02

sec2α

⇒ sinβ = cosβtanα � grcos2β

2v02cos2α

⇒ grcos2β

2v02cos2α

= cosβtanα � sinβ

⇒ grcos2β

2v02cos2α

= cosβsinα

cosα� sinβ

⇒ grcos2β

2v02cos2α

= sinαcosβ � cosαsinβ

cosα

⇒ grcos2β

2v02cosα

= sinα � β�

⇒ r = 2v02

gcos2βcosαsinα � β� ____________(i)

Since sin(α + β) � sin(α � β) = 2cosαsinβ

Replacing β by α � β, we get

sin(2α � β) � sin(β) = 2cosαsin(α � β) ___________(ii)


r = v0

2

g

sin(2α � β) � sinβ

cos2β

Which is the range of a projectile on an inclined plane.

Range will maximum if sin(2α � β) = 1

⇒ 2α � β = sin – 1(1)

⇒ 2α = β + 900

⇒ α = β

2 + 45

0

Thus,

r<=> = v0

2

g�1 � sinβ

cos2β�

= v0

2

g� 1 � sinβ

1 � sin2β

�

= v0

2

g� 1 � sinβ1 � sinβ�1 + sinβ�� =

v02

g1 + sinβ�


30

� Question 17

A fort and a ship are both armed with guns which give their projectiles a muzzle velocity �2gk, and the guns in the fort are at a height h above the guns in the ship. If d1 and d2 are the greatest horizontal ranges at which the fort and ship, respectively, can engage, prove that

d1

d2=�k + h

k � h

Solution F r

h

v0

α β

x

S d2 A

Let S be ship and F be fort. Let fort makes angle β with x-axis. i.e. ∠ASF = β. Let SF = r Since d2 is greatest horizontal range for gun in the ship so r is the maximum range on inclined

plane with inclination β. Then

r = v0

2

g1 + sinβ� Put v0 = �2gk

⇒ r = 2gk

g1 + sinβ� ⇒ 2k r + rsinβ

From fig.

h = rsinβ

⇒ 2k r + h

⇒ r 2k – h

Also from fig.

r2= h2+ d2

2


31

⇒ 2k – h�2 = h2+ d2

2

⇒ 4k2+ h

2 � 4hk = h2+ d2

2

⇒ d22 4k

2 � 4hk 4kk � h� _______(i)

For a gun in fort, change h to – h

d12 4kk + h� _______(ii)


d12

d22

4kk + h�4kk � h�

⇒ d1

d2

�k + hk � h

� Question 18

A shell of mass m1 + m2 is fired with a velocity whose horizontal and vertical components are u, v and at the highest point in its path the shell explodes into two fragments m1, m2. The explosion produces additional kinetic energy E, and the fragments separate in a horizontal direction. Show that they strike the ground at a distance apart which is equal to

V

g�2E � 1

m1

� 1

m2

�

Solution

Let v0 be the velocity of the projection. then by given conditions

u = v0cosα _______(i)

v = v0sinα ______(ii)

At the highest point, there is only horizontal velocity u. Let v1 and v2 be the velocities of m1 and m2 respectively at the time of explosion. Then by law of conservation of momentum.

m1v1 + m2v2 = (m1 + m2)u

⇒ u = m1v1 + m2v2

m1 + m2

_______(iii)

Now Increase in K.E. = K.E. after explosion – K.E. before explosion

⇒ E = �1

2m1v1

2 + 1

2m2v2

2� � �1

2m1+ m2�u2�

⇒ 2E = m1v12 + m2v2

2 � m1+ m2�u2


32

⇒ 2E = m1v12 + m2v2

2 � m1+ m2� �m1v1 + m2v2

m1 + m2

�2 By (iii) ⇒ 2m1+ m2�E = m1+ m2� m1v1

2 + m2v22� � m1v1 + m2v2�2 = m1

2v12 + m1m2v2

2 + m2m1v12 + m2

2v22 � m1

2v12 � m2

2v22 � 2m1m2v1v2 = m1m2v2

2 + m2m1v12 � 2m1m2v1v2 = m1m2v2

2 + v12 � 2v1v2� = m1m2v2 � v1�3

⇒ v2 � v1�3 2m1+ m2�E

m1m2

⇒ v2 � v1 �2E � 1

m1

� 1

m2

�

Which is relative velocity of m1 + m2.

Now the time taken by pieces to touch the ground is given by:

Time = 1

2(Time of flight)

= 1

2�2v0sinα

g� =

v0sinα

g

= v

g By (ii)

Distance Apart = (Relative velocity)(Time)

v2 � v1� v

g

v

g�2E � 1

m1

� 1

m2

�

Which is required.


33

� Question 19 (Speed of projectile) Show that least speed with which a particle must be projected so that it passes through two points P and Q at height hP and hQ respectively is:

�g�hP + hQ+ PQ�

Solution y M N Directrix

y = v02

2g

P

Q

hP

S hQ

x O L R Let S be the focus.

From fig.

LM = LP + PM

⇒ v0

2

2g = hP+ PM _________(i)

RN = RQ + QN

⇒ v0

2

2g = hQ+ QN _________(ii)

Adding (i) and (ii), we get

v02

g = hP+ hQ+ PM + QN

⇒ v02 = g�hP+ hQ+ PM + QN�

= g�hP+ hQ+ PS + QS� By focus-directrix property


34

⇒ v0 = �g�hP+ hQ+ PS + QS�

v0 is least when PS + QS is least, which is least when S lies on PQ. i.e. when

PS + QS = PQ

Hence

v0 �min = �g�hP+ hQ+ PQ�

%%%%% End of The Chapter # 7 %%%%%

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