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Projectiles
In the last chapter we considered the motion of a particle along a straight line. But we observe that a particle moves along a curved path if it is freely projected in the air in the direction other than vertical. These freely projected particles which are having the combined effect of a vertical and a horizontal motion are called projectiles. The motion of a projectile has a vertical component and a horizontal component. The vertical component of the motion is subjected to gravitational acceleration/retardation while horizontal component remains con- stant, if air resistance is neglected. The motion of a projectile can be analysed independently in vertical and horizontal directions and then combined suitably to get the total effect. In this chapter, after defining important terms to be used, horizontal projection, inclined projection on both horizontal and inclined plane are analysed neglecting air resistance.
13.1. DEFINITIONS
The definitions of the terms used in this chapter are given with reference to Fig. 13.1. Velocity of projection : The velocity with which the particle is projected is called as velocity of projection ( u m/sec). Angle of projection : The angle between the direction of projection and horizontal direction is called as angle of projection (α). Trajectory : The path traced by the projec- tile is called as its trajectory. Horizontal range : The horizontal distance through which the projectile travles in its flight is called the horizontal range or simply range of the projectile. Time of flight : The time interval during which the projectile is in motion is called the time of flight.
13.2. MOTION OF BODY PROJECTED HORIZONTALLY
Consider a particle thrown horizontally from point A with a velocity u m/sec as shown in Fig. 13.2. At any instant the particle is subjected to: (1) Horizontal motion with constant velocity u m/sec. (2) Vertical motion with initial velocity zero and moving with acceleration due to gravity g.
Fig. 13.
Fig. 13.
y u
Trajectory
A a RangeRange x
hh
A (^) u
u gt
RangeRange B
324 FUNDAMENTALS OF ENGINEERING MECHANICS
Let h be the height of A from the ground. Considering vertical motion,
h = 0 × t +
gt^2
gt^2 ...(13.1)
This expression gives the time of flight. During this period, the particle moves horizontally with uniform velocity, u m/sec. ∴ Range = ut …(13.2) Example 13.1. A pilot fying his bomber at a height of 2000 m with a uniform horizontal velocity of 600 kmph wants to strike a target (Ref : Fig. 13.3). At what distance from the target, he should release the bomb? Solution. h = 2000 m u = 600 kmph
= 600 1000 60 60
×
×
= 166.67 m/sec Initial velocity in vertical direction = 0 and gravitational acceleration = 9.81 m/sec 2. If t is the time of flight, considering vertical motion, we get
2000 = 0 × t +
2 × 9.81^ t
2
∴ t = 20.19 sec During this period horizontal distance travelled by the bomb = ut = 166.67 × 20. = 3365.46 m Bomb should be released at 3365.46 m from the target. Ans. Example 13.2 A person wants to jump over a ditch as shown in Fig. 13.4. Find the minimum velocity with which he should jump. Solution. h = 2 m and Range = 3 m Let t be the time of flight and u the minimum horizontal velocity required. Considering the verti- cal motion :
h = 1 2
gt^2
9.81 t^2
∴ t = 0.6386 sec
Fig. 13.
Fig. 13.
hh = 2000 m = 2000 m
A (^) u = 600 kmph
R = ?R =? B
A
u
3m3m
2m
B
326 FUNDAMENTALS OF ENGINEERING MECHANICS
In the horizontal direction, rocket has initial velocity = 1200 kmph and acceleration = 6 m/sec 2 Now, u = 1200 kmph
=
×
×
= 333.33 m/sec
∴ Horizontal distance covered during the time of flight = range
= ut +
at^2
= 333.33 × 24.73 +
× 6 × 24.73 2
= 10,078.5 m ∴ The angle θ below the horizontal at which the pilot must see the target while releasing the rocket, is given by
tan θ = 3000 10 078 5,. ∴ θ = 16.576° Ans.
13.3. INCLINED PROJECTION ON LEVEL GROUND
Consider the motion of a projectile, projected from point A with velocity of projection u and angle of projection α, as shown in Fig. 13.7. Let the ground be a horizontal surface. The particle has motion in verti- cal as well as horizontal directions. Vertical Motion Initial velocity = u sin α upward Gravitational acceleration = g = 9.81 m/sec 2 downward i.e. a = – g = – 9.81 m/sec 2 Hence initially the particle moves upward with velocity u sin α and retardation 9.81 m/sec^2. Velocity becomes zero after some time (at C ) and then the particle starts moving downward with gravitational acceleration. Horizontal Motion Horizontal component of velocity = u cos α. Neglecting air resistance, we can say that the projectile is having uniform velocity u cos α during its entire flight. Equation of the Trajectory Let P ( x, y ) represent the position of projectile after t seconds. Considering the vertical motion,
y = ( u sin α) t – 1 2
gt^2 …(13.3)
Fig. 13.
y u
a
P(x, y)
C
A (^) R = RangeR = Range
Maximum height
Maximum height
B x
PROJECTILES 327
Considering horizontal motion, x = ( u cos α) t …(13.4)
∴ t = x u cos α Substituting this value in Eqn. 13.3, we get
y = u sin α x u
g x cos α u cosα
F
HG^
I
KJ
2
i.e. y = x tan α –
2 2 2
gx u cos α
But (^12) cos α
= sec 2 α = (1 + tan^2 α).
Hence Eqn. (13.5) reduces to the form
y = x tan α –
2 2
x u
(1 + tan^2 α) …(13.6)
This is an equation of a parabola. Hence the equation of the trajectory is a parabola. Maximum Height When the particle reaches maximum height, the vertical component of the velocity will be zero. Considering vertical motion, initial velocity = u sin α final velocity = 0 acceleration = – g Using the equation of linear motion v^2 – u^2 = 2 a s , we get 0 – ( u sin α) 2 = – 2 gh
h = u g
2 2 2
sin α (^) …(13.7)
Time Required to Reach Maximum Height Using first equation of motion ( v = u + at ), when projectile reaches maximum height, 0 = u sin α – gt
∴ t = u g
sin α (^) …(13.8)
Motion of the projectile in verticle direction is given by Eqn. 13.3 as
y = ( u sin α) t – 1 2
gt^2 At the end of flight, y = 0
∴ 0 = ( u sin α) t – 1 2
gt^2
= t^ F u sin α − gt
HG^
I
KJ
PROJECTILES 329
Example 13.5. A body is projected at an angle such that its horizontal range is 3 times the maximum height. Find the angle of projection. Solution. Let u be velocity of projection and α the angle of projection. Then
Maximum height reached = u g
2 2 2
sin α
and Range = u g
(^2) sin 2 α
∴ In this case u g
u g
sin α sin α = ×
∴ sin 2α = 3 2
sin 2 α
i.e. 2 sin α cos α =
sin^2 α
or tan α = 4 3 i.e. α = 53.13 degree Ans. Example 13.6. A projectile is aimed at a target on the horizontal plane and falls 12 m short when the angle of projection is 15°, while it overshoots by 24 m when the angle is 45°. Find the, angle of projection to hit the target. Solution. Let s be the distance of the target from the point of projection and u be the velocity of projection, as shown in Fig. 13.9. Range of projection is given by the expression
R = u g
(^2) sin 2 α
Applying it to first case,
s – 12 = u g
2 sin (2 × 15°)
= u g
u g
× = …(1)
From the second case
s + 24 + = u g
2 sin (2 × 45°)
u g
2 …(2)
Fig. 13.
ss – 12 – 12 ss s + 24s + 24
330 FUNDAMENTALS OF ENGINEERING MECHANICS
From (1) and (2), we get s + 24 = 2 ( s – 12) ∴ s = 48 m Let the correct angle of projection be α. Then
48 = u g
2 sin 2α …(3)
From eqn. (2) u g
2 = s + 24 = 48 + 24
= 72m ∴ From eqn. (3) 48 = 72 sin 2α ∴ 2 α = 41.81° α = 20.905° Ans. Example 13.7. The horizontal component of the velocity of a projectile is twice its initial vertical component. Find the range on the horizontal plane, if the projectile passes through a point 18 m horizontally and 3 m vertically above the point of projection. Solution. Let u be the initial velocity and α its angle of projection. Vertical component of velocity = u sin α Horizontal component of velocity = u cos α In this problem, u cos α = 2 u sin α
∴ tan α =
∴ α = 26.565° It is given that when x = 18 m, y = 3m. Using the equation of trajectory
y = x tan α –
gx u
2 (^2) cos 2 α
we get, 3 = 18 × 1 2
2 − × (^2 )
×
u cos.
[since tan α = 1/2]
i.e. u^2 = 9 81^18 6 2 26 565
2 2
cos.
×
× ×
i.e. u = 18.196 m/sec ∴ Range on the horizontal plane
= u g
(^2) sin 2 α
= 18 196^2 26 565
. 2 sin (. ) .
× × °
= 27.00 m Ans.
332 FUNDAMENTALS OF ENGINEERING MECHANICS
Any one of the following three methods can be used to analyse such cases :
Method 1 From equation of motion in verti- cal direction,
y = u sin α × t – 1 2
gt^2
By putting y = – y 0 in the above equation, the time required to reach B (time of flight) is obtained.
gt^2
Once the time of flight is known, horizontal range can be found from the relation : R = u cos α × t Maximum height above the point of projection and time required to reach it can be found as usual from Eqn. (13.7) and (13.8)
Method 2 Total time of flight can be split into two parts t 1 , time required to reach maximum height (point C ) and t 2 , time required to descend from point C to B. Considering the vertical motion and noting that vertical component of velocity is zero at C , we get, 0 = u sin α – gt 1
t 1 = u g
sin α
Height reached in this time interval,
h = u sin α × t 1 – 1 2
gt 12
u g
g u g
2 2 2 2 2
sin α sin α −
= u g
2 2 2
sin α
While descending, the vertical distance to be covered = h + y 0. Considering downward motion
h + y 0 = 0 × t 2 +
gt 22 Hence t 2 can be found. Therefore total time of flight t = t 1 + t 2. Horizontal range is given by R = u cos α × t
Fig. 13.
y
u
a
C
A yoyo
x
B
PROJECTILES 333
Method 3 The motion can be split into two parts ; AD and DB where D is the point on the trajec- tory at the same level as A , the point of projec- tion (See Fig. 13.12). For the portion AD , the Eqn. (13.7) to (13.10) can be used. For the portion DB : The vertical component of the velocity of the projectile at point D will be equal to the vertical component of the velocity at A , but downwards, since A and D are at the same eleva- tion. The horizontal component of the velocity remains equal to u cos α through out. Hence for vertical motion in the portion DB initial velocity = u sin α, downward gravitational aceleration = g = 9.81 m/sec 2 downward.
∴ y 0 = ut 2 sin α + 1 2
gt 22
where, t 2 is the time taken to travel portion DB. Horizontal distance moved during this time = u cos α × t 2 Total result may be obtained by combining the motion in the two portions. Example 13.9. A bullet is fired from a height of 120 m at a velocity of 360 kmph at an angle of 30° upwards. Neglecting air resistance, find ( a ) total time of flight, ( b ) horizontal range of the bullet, ( c ) maximum height reached by the bullet, and ( d ) final velocity of the bullet just before touching the ground. Solution. Velocity of projection u = 360 kmph
= 360 1000 60 60
×
×
= 100 m/sec.
( a ) Total time of flight Total time of flight can be found from any one of the three methods. Here all the three methods are illus- trated referring to Fig. 13.13. Method 1 y 0 = – 120 m. Considering vertical motion
y = u sin α × t –
gt^2 ,
× 9.81 t^2
Fig. 13.
Fig. 13.
A a
u C
D u cos^ a
u sin a B
y
A
yy (^) o (^) o= 120 m= 120 m
360 kmph C D (^) x
B
30°
PROJECTILES 335
( c ) Horizontal range = u cos α × t = 100 cos 30° × 12. = 1056.55 m Ans. ( d ) Velocity of the bullet just before striking the ground : Vertical component of velocity = u sin α – gt = 100 sin 30° – 9.81 × 12. = – 69.682 m/sec = 69.682 m/sec downward. Horizontal component of velocity, = 100 cos α = 86.603 m/sec Referring to Fig. 13.14, the velocity at strike v = (^) 69 682. 2 +86 603.^2 v = 111.16 m/sec Ans.
θ = tan–^
= 38.82° as shown in Fig. 13.14 Ans. Example 13.10. A cricket ball thrown by a fielder from a height of 2m, at an angle of 30° to the horizontal, with an initial velocity of 20 m / sec, hits the wickets at a height of 0.5 m from the ground. How far was the fielder from the wickets? Solution. Initial velocity u = 20 m/sec. Angle of projection α = 30° y 0 = – (2.0 – 0.5) = – 1.5 m Time of flight t is given by the expression
× 9.81 t^2 t^2 – 2.0387 t – 0.3058 = 0 i.e. t = 2.179 sec. ∴ The distance of the fielder from the wickets = Range = u cos α × t = 20 cos 30° × 2. = 37.742 m Ans. Example 13.11. Gravel is thrown into a bin from the top of a conveyor ( Ref. Fig. 13.16 ) with a velocity of 5 m / sec. Determine : ( a ) time it takes the gravel to hit the bottom of the bin ; ( b ) the horizontal distance from the end of the conveyor to the bin where the gravel strikes the bin, and ( c ) the velocity at which the gravel strikes the bin.
Fig. 13.
Fig. 13.
q
30°
20 m/sec
2m2m 0.5 m
336 FUNDAMENTALS OF ENGINEERING MECHANICS
y
A x
50°
10m10m
B
Fig. 13. Solution. Taking A as origin of the coordinate system as shown in the figure, vertical motion is represented by :
y = u sin α × t – 1 2
gt^2
where, u = 5 m/sec and α = 50° For the point B ; y = – 10 m
∴ – 10 = 5 sin 50° × t – 1 2
9.81 t^2
i.e. t = 1.871 sec Ans. Horizontal distance travelled in this time = u cos α × t = 5 cos 50° × 1. = 6.012 m Ans. Vertical component of velocity of gravel at the time of striking the bin is given by : = u sin α – gt = 5 sin 50° – 9.81 × 1. = – 14. = 14.524 m/sec (downward). Horizontal component of velocity = u cos α = 5 cos 50° = 3.214 m/sec Velocity of strike
v = 14 524.^2 +3 214.^2 = 14.875 m/sec Ans.
θ = tan –1^ 14 524 3 214
= 77.52° to horizontal as shown in Fig. 13.17 Ans.
Fig. 13.
q
v
338 FUNDAMENTALS OF ENGINEERING MECHANICS
v (^) y = 42.02 m/sec
v = v (^) x^2^ + vy^2 = 14 007. 2 +42 02.^2 = 44.294 m/sec Ans.
θ = tan –^ v v
y x
F
HG^
I
KJ^
= 71.565° to horizontal Ans. Example 13.13. A ball rebounds at A and strikes the inclined plane at point B at a distance s = 76 m as shown in Fig. 13.19. If the ball rises to a maximum height h = 19 m above the point of projection, compute the initial velocity and the angle of projection α. Solution. At A the vertical component of velocity = u sin α When maximum height h = 19 m is reached, vertical component of velocity is zero. ∴ 0 – ( u sin α) 2 = 2 (– g ) 19 Substituting g = 9.81 m/sec^2 u sin α = 19.308 …(1) y -co-ordinate of B = – AB × sin θ
= – 76 × 1 3 2 + 12
= – 24.033 m
Considering the motion in vertical upward direction,
9.81 t^2
= 19.308 t – 1 2
9.81 t^2
t^2 – 3.936 t – 4.9 = 0 ∴ t = 4.93 sec. x co-ordinate of B = AB × cos θ
= 76 ×
= 72.1 m
Considering the horizontal motion of the ball u cos α × t = 72. u cos α × 4.93 = 72. or u cos α = 14.625 …(2) From eqn. (1) and (2)
tan α =
Fig. 13.
y
A
a hh = 19 m^ =^19 m
3 1
q
x
B
S = 76 mS = 76 m
PROJECTILES 339
∴ α = 52.86° Ans. From Eqn. (2),
u =
cos. °
i.e. u = 24.222 m/sec Ans. Note : Though this example is about projection on inclined plane, to be discussed in next article, it could be solved now only as x , and y -coordinates of the point of strike B are known. Example 13.14. A bomber is flying horizontally at an altitude of 2400 m with the uniform velocity of 1000 kmph to bomb a target. Where should the bomb be released to strike the target? At the same time, the target is being defended by an anti-aircraft gun at 60° to the hori- zontal. The muzzle velocity of the gun is 600 m / sec. When should the shell be fired to hit the bomber? Solution. Altitude of the bomber h = 2400 m. Let the time required for bomb to reach ground be t seconds. Then, since initial vertical velocity = 0,
2400 = 0 × t +
× 9.81 t^2 t = 22.12 sec
Now, u = 1000 kmph =
×
×
= 277.78 m/sec. Horizontal distance moved by bomb = u × t = 277.78 × 22. = 6144.493 m. Bomb should be released when the bomber is 6144.493 m away from the target. Ans.
2400
30658.58 m 33111 m (^) 1233.07 m 1331.7 m 6144.501 m
Fig. 13. Muzzle velocity of the gun = 600 m/sec i.e. the velocity of projection u = 600 m/sec Anlge of projection α = 60°
PROJECTILES 341
y
A b
a
RR (^) B
D (^) x
Fig. 13. The equation of trajectory of the projectile is given by
y = x tan α – 1 2
2 2 2
gx u cos α Applying this equation to point B , we get
R sin β = R cos β tan α – 1 2
g R u
2 2 2 2
cos cos
β α
i.e. R
2 2 2
g u
cos cos
β α
F
HG^
I
KJ^
= cos β tan α – sin β
∴ R =
2
u g
cos cos
α β
(cos β tan α – sin β)
2
u g
cos cos
α β
(cos β. sin α – sin β. cos α)
2
u g
cos cos
α β
sin (α – β) …(13.13( a ))
i.e. R = u g
2 cos^2 β [sin (2α^ –^ β) – sin^ β]^ …(13.13 ( b )) [since 2 cos A sin B = sin ( A + B ) – sin ( A – B )] Time of flight : Let t be the time of flight. The horizontal distance covered during the flight = AD = u cos α × t
∴ t = AD u cos α
R
u
cos cos
β α
2
u g u
cos cos
sin ( ) cos
α β
α β α
× −^ cos β
2 u g
sin ( ) cos
α β β
342 FUNDAMENTALS OF ENGINEERING MECHANICS
For the given values of u and β, the range is maximum when : sin (2α – β) = 1
i.e. 2 α – β = π 2
or α = π^ β 4 2
Referring to Fig. 13.
θ 1 = π^ β 4 2
π β 4 2
and θ 2 = π^ β 2
Thus θ 2 = 2θ 1 i.e. the range on the given plane is maximum, when the angle of projection bisects the angle between the vertical and inclined planes.
If the projection is down the plane, the Fig. 13.13 to 13.15 can be still used, but the value of β should be taken negative.
Example 13.15. A plane has a slope of 5 in 12. A shot is projected with a velocity of 200 m / sec at an upward angle of 30° to horizontal. Find the range on the plane if :
( a ) the shot is fired up the plane ; ( b ) the shot is fired down the plane. Solution. Initial velocity u = 200 m/sec. Angle of projection α = 30°
Inclination of the plane = tan –^
F
HG^
I
KJ^ = 22.62°
( a ) When the shot is fired up the plane : β = + 22.62°
Range = u g
2 cos 2 β [sin (2α^ –^ β) – sin^ β]
2
. × cos^2. [sin (2 × 30 – 22.62) – sin 22.62°]
i.e. Range = 1064.65 m Ans.
( b ) When the shot is fired down the plane : β = – 22.62°
Fig. 13.
A (^) b
a
q 1
q 2
Horizontal
B
Direction for maximum projection
Vertical
