An Analysis of Projectile Motion Projectile Motion Lab Report
PROJECTILE MOTION - Weebly
Transcript of PROJECTILE MOTION - Weebly
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M th
� 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
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⇒ 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
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⇒ 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
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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�
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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
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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
R will be maximum when sin2α is maximum.
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
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We know that
Rage = (Horizontal Velocity)(Time of Fight)
v0cosα� �2v0sinα
g� v0
2
gsin2α
R will be maximum when sin2α is maximum.
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
As Height reached v02sin2α
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
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We know that
Rage = (Horizontal Velocity)(Time of Fight)
v0cosα� �2v0sinα
g� v0
2
gsin2α
R will be maximum when sin2α is maximum.
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
As Height reached v02sin2α
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)
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v0cosα� �2v0sinα
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
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We know that
Rage = (Horizontal Velocity)(Time of Fight)
⇒ 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
Rage = (Horizontal Velocity)(Time of Fight)
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⇒ 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
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� 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'
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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
Rage = (Horizontal Velocity)(Time of Fight)
v0cosα� �2v0sinα
g�
v02
gsin2α
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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.
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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).
We know that the Cartesian equation of the trajectory of a projectile is:
y = xtanα � gx2
2v02
sec2α
Since P(a, � h) lies on it, therefore
� h = atanα � ga2
2v02
sec2α
⇒ � 2v02h = 2av0
2tanα � ga2�1 & tan2α�
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⇒ 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)
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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)
We know that the Cartesian equation of the trajectory of a projectile is:
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 + Dtan�
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
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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
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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.
We know that the Cartesian equation of the trajectory of a projectile is:
y = xtanα � gx2
2v02
sec2α
Putting y = h and v0 = �2gH, we get
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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
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� 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
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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.
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� 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
y = (v0sinα)t � 1
2gt2
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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
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� 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
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⇒ 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)
y = (v0sinα)t � 1
2gt2 ______(iii)
At t = 2a
V ii� becomes
x = (v0cosα)2a
V
= (v0cosα)2a
v0sinα
⇒ x = 2acotα ______(iv)
At t = 2a
V iii� becomes
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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
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= 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
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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)
From (i) and (ii), we get
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�
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� 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
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⇒ 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)
From (i) and (ii), we get
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
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⇒ 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.
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� 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
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⇒ 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 %%%%%