Available At: mathcity.org Contact At: [email protected]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 ‘v 0 ’ 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) v 0 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 dt i + dy dt j ⇒ v = dx dt i + dy dt j CHAPTER 7 PROJECTILE MOTION
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PROJECTILE MOTION - MathCity.org PROJECTILE MOTION 7 . Available At: mathcity.org Contact At: [email protected] 2 ⇒ dv dt = d2x 2 i + d2y dt2 j ⇒ a = d2x dt2 i + d2y dt2 j ...
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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.
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.
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
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
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
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.
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
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
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:
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
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
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:
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
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.
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.
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)
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
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
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
� 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: