Chapter no. 7 projectile

Chapter No. 7 Circular Motion

CHAPTER NO.7

CIRCULAR MOTION

7.1 Projectile Motion (an example of circular motion)

Terms:

7.1.1 Projectile (Nothing but particle):

It is a particle or object projected in the space at an inclination to the direction of gravity, moving

under the combined effect of vertical and horizontal forces.

1) Trajectory:

It is a path, traced by a projectile in the space.

2) Point of Projection:

It is the initial point of the particle (see Fig. 1 Point A)

3) Angle of inclination or Projection:

The angle, with a horizontal, at which a projectile is projected (see Fig. 1)

4) Horizontal Range or Range (R) :

The distance between the point of projection and the point where projectile strikes the

ground (see Fig. 1)

5) Time of Flight:

The total time taken by the projectile, to reach the maximum height i.e. hmax and to return

back to the ground.

6) Velocity of projection:

The velocity with which, a projectile is projected (see Fig. 1)


Fig. 1

7.2 Study of a projectile:

1) Generally, the velocity of the projection ‘u’ split up in to or resolved into two

components ‘u cos α’ and ‘u sin α’.

i. e. Vertical component V = u sin α

and Horizontal component H = u cos α

2) The component in vertical direction is subjected to retardation due to gravity. Therefore,

this vertical velocity ‘u sin α’ is subjected to retardation for first half of trajectory before

attaining maximum height i.e. hmax and for second half it is subjected to acceleration due

to gravity, before coming to rest.

3) The component in horizontal direction, there is no retardation or acceleration due to

gravity. Therefore, this horizontal velocity ‘u cos α’ is remain constant throughout the

motion.

For first half of trajectory ---------- {

𝑣 = 𝑢 sin 𝛼 − 𝑔𝑡

𝑣2 = 𝑢2 sin2 𝛼 − 2𝑔𝑠

𝑠 = ( 𝑢 sin 𝛼 ) 𝑡 − (1

2 𝑔 𝑡2)


For second half of trajectory ---------- {

𝑣 = 𝑢 sin 𝛼 + 𝑔𝑡

𝑣2 = 𝑢2 sin2 𝛼 + 2𝑔𝑠

𝑠 = ( 𝑢 sin 𝛼 ) 𝑡 + (1

2 𝑔 𝑡2)

4) Thus, the time taken by the body or particle or projectile to reach the ground, is

calculated by vertical component (u sin α ) of the velocity and the horizontal component

of the velocity ( u cos α).

5) Mathematically,

[𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑅𝑎𝑛𝑔𝑒 (𝑅)

𝑜𝑟 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒

] = [𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

(𝑢 cos 𝛼 )] 𝑥 [

𝑇𝑖𝑚𝑒(𝑡) ]

Fig. 2

7.3 Derivation of the expression or equation for the path of a projectile:

Consider a particle projected upward from point A, at an angle α, with the horizontal, with an

initial velocity u m/s as shown in Fig. 1 below.

Now, the velocity of projection u can be split up into two components i.e. u cos α and u sin α.


Fig. 3

Consider any point as the position of the particle, after time t seconds with co-ordinates x and y

as shown in Fig. 1

𝑦 = ( 𝑢 sin 𝛼 ) 𝑡 − (1

2 𝑔 𝑡2) ------------------- (1)

And 𝑥 = ( 𝑢 cos 𝛼 ) 𝑡 ----------- (2)

Therefore, 𝑡 = 𝑥

𝑢 cos 𝛼

Substituting in (1)

𝑦 = ( 𝑢 sin 𝛼 ) 𝑥

𝑢 cos 𝛼− (

1

2 𝑔

𝑥2

𝑢2 cos2 𝛼)

𝑦 = 𝑥 tan 𝛼 − 𝑔 𝑥2

2 𝑐𝑜𝑠2 𝑢2 ---------- (3)

Equation (3) is known as equation of trajectory

7.4 Derivation of expression for the time of flight of the projectile on a horizontal plane:

We have, 𝑠 = 𝑢𝑡 + 1

2 𝑎 𝑡2


𝑦 = ( 𝑢 sin 𝛼 ) 𝑡 − (1

2 𝑔 𝑡2) ------------------- (1)

i.e. s = y, u = u sin α and a = g

When the projectile p is at point B, y = 0

Substituting, the value of y in eqn (1)

0 = ( 𝑢 sin 𝛼 ) 𝑡 − (1

2 𝑔 𝑡2)

( 𝑢 sin 𝛼 ) 𝑡 = (1

2 𝑔 𝑡2)

( 𝑢 sin 𝛼 ) = (1

2 𝑔 𝑡)

𝑡 = 2 𝑢 sin 𝛼

𝑔

i.e. 1) time required to reach the maximum height

𝑡

2=

𝑢 sin 𝛼

𝑔


7.5 Derivation of the expression for horizontal range (R) of the projectile:

We know,

Horizontal component of velocity = u cos α

And

Time of flight,

𝑡 = 2 𝑢 sin 𝛼

𝑔

Therefore,

[Horizontal range ] = [𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦] 𝑥 [Time of flight]

= [u cos α] 𝑥 [2 𝑢 sin 𝛼

𝑔]

𝑅 = sin 2 𝛼 𝑢2

𝑔

Note:

The range will be maximum when sin 2α = 1


2α = 900

α = 450

Rmax = 𝑢2

𝑔

7.6 Derivation of expression for the maximum height of a projectile on a horizontal plane:

We have, the vertical component of initial velocity u is

v = u sin α

and vertical component of final velocity

v = 0

Therefore,

Average velocity = 𝑢 sin 𝛼+0

2

= 𝑢 sin 𝛼

2

Now,

[𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒] = [𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦] x [𝑡𝑖𝑚𝑒]


= 𝑢 sin 𝛼

2 𝑥

𝑢 sin 𝛼

𝑔

= 𝑠𝑖𝑛2 𝛼 𝑢2

2 𝑔

Where, 𝑢 sin 𝛼

𝑔 = is the time required by the projectile to reach the maximum height

H = 𝑠𝑖𝑛2 𝛼 𝑢2

2 𝑔

Questions:-

1. A projectile is fired upwards at an angle of 300 with a velocity of 40 m/s. Calculate the

time taken by the projectile to reach the ground after firing.

2. If a particle is projected inside a horizontal tunnel which is 5 m high with a velocity of

60 m/s, find the angle of projection and the greatest possible range.

3. The range of projectile on a horizontal plane is 240 m and the time of flight is 8 sec. Find

the initial velocity and the angle of projection.

4. A body is projected at such an angle that the horizontal range is three times the greatest

height. Find the angle of projection.

5. Find the least initial velocity which a projectile may have, so that it may clear a wall 3.6

m high and 4.8 m distant (from the point of projection) and strike the horizontal plane

through the foot of the wall at a distance 3.6 m beyond the wall. The point of projection is

at the same level as the wall.

6. Find the angle of projection at which the horizontal range and the maximum height of a

projectile are equal. May 2010 (06 MKS)

7. A particle is projected in air with a velocity 100 m/s and at an angle of 300 with the

horizontal. Find:

1) The horizontal range

2) The maximum height by the particle

3) And the time of flight

Ans: R = 882.8 m, Hmax = 127.42 m, t = 10.19 sec


8. A projectile is fired with an initial velocity of 250 m/s at a target located at a horizontal

distance of 4 km and the vertical distance of 700 m above the gun. Determine the value of

firing angle to hit the target. (Neglect air resistance).

Ans: α = 68.820 and 31.080

9. A projectile is aimed to a mark on a horizontal plane through the point of projection and

fall 12 m short of the mark and with an angle of projection 150. While it overshoots the

mark by 24 m when the angle of projection to hit the mark. Assume no air resistance.

Take the velocity of projection is constant in all cases.

Ans: R = 48 m, u = 26.57 m, α = 20.550

Case I) Projectile projected horizontally from certain height

a) Considering vertical motion of projectile

We have,

S = 𝐮𝐭 +𝟏

𝟐 𝐚𝐭𝟐

Here,

S = y = vertically downward displacement

u = initial velocity = u sin α = 0

a = acceleration due to gravity = g = 9.81 m/s2

∴ y = 𝟏

𝟐 𝐠 𝐭𝟐---------- (1)

b) Considering horizontal motion of projectile


x = u x t --------- (2)


Problems:

1) An aircraft moving horizontally at a speed of 1000 kmph at a height of 1500 m releases a

bomb which hits the target.

Find:

1) Time required for the bomb to reach the target on ground

2) The horizontal distance of the aircraft from the target when it releases the

bomb. (May 2007 8 MKS)

Ans: t = 17.49 sec and x = 4857.66 m

2) Solve the same problem when y = 2000 m and u = 150 m/s.

Ans: t = 20.2 sec and x = 3030m

3) Find out u = ? when y = 100 m and x = 200 m

Ans: t = 4.515 sec and u = 44.3 m/s

4) Find y = ? and u = ? when time required to strike the ground is 15 sec when x = 50 m.

Ans: y = 1103.625 m and u = 3.33 m/s

5) A person wants to jump over a ditch as shown in Fig. below. Find the max. velocity with

which he should jump? (May 2006 6 MKS)

Ans: t = 0.638 sec and u = 4.702 m/s


Case II) Projectile projected at an angle α from certain height (say y)

a) Considering vertical motion of projectile

We have,

S = 𝐮𝐭 +𝟏

𝟐 𝐚𝐭𝟐

Here,

S = y = vertically downward displacement

u = initial vertical velocity = u sin α = 0

a = acceleration due to gravity = g = 9.81 m/s2

∴ − y = (u sin α) – 𝟏

𝟐 𝐠 𝐭𝟐

y = - (u sin α) + 𝟏

𝟐 𝐠 𝐭𝟐---------- (1)

b) Considering horizontal motion of projectile


x = u cos α x t --------- (2)

Velocity with which it strikes the ground (VB):

VB = u = initial velocity


VH = u cos α

VV = u sin α

We have,

𝐯 = 𝐮 + 𝐚𝐭

-VV = u sin α – gt

VV = -u sin α + gt

VB = √ 𝑽𝑯 𝟐 + 𝑽𝑽

𝟐--------------- (velocity of hit)

Direction of hit:

θ = tan-1 ( 𝑽𝑽

𝑽𝑯 ) ------------ (with the horizontal)

Problems:

1. A body is projected from the top of the tower 40 m high and strikes the ground after 10

seconds at a point 400 m from the base of the tower. Determine the velocity and angle of

projection. Also determine the maximum height attained by the body above ground level.

Dec. 2009 (10 MKS)

Ans: u = 60.24 m/s and Hmax = 143.43 m

2. A cricket ball is thrown by a fielder from a height of 2 , at an angle of elevation 300 to the

horizontal with an initial velocity of 20 m/s, hits the wickets at a height of 0.5 m from the

ground. How far the fielder from the wicket? (May 2004 12 MKS)

Ans: t = 2.181 sec, x = 37.77 m

3. A solider fires a bullet at an angle of elevation 300 from his position on the top of hill to

strike a target which is 60 m lower than the position of the solider. The initial velocity of

the bullet is 75 m/s. Calculate:

a) Hmax to which bullet rise above horizontal

b) The actual velocity with which it will strike the target

c) Total time required by for the flight of the bullet


Ans: Hmax = 131.74 m, t = 9.01 sec, vB = 82.456 m/s, θ = 38.020

4. A projectile fired from the edge of a 150 m high cliff with an initial velocity of 180 m/s at

an angle of elevation of 300 with the horizontal. Neglecting air resistance. Find:

1) The greatest elevation above the ground reached by the projectile and

2) Horizontal distance from gun to the point, where the projectile strikes the ground.

Ans: Hmax = 563.3 m, t = 19.9 sec, R = 3.012 km

Case III) Projectile projected at an angle α from certain height to the downward (say y)

c) Considering vertical motion of projectile

We have,

S = 𝐮𝐭 +𝟏

𝟐 𝐚𝐭𝟐

Here,

S = - y = vertically downward displacement

u = initial vertical velocity = - u sin α

a = acceleration due to gravity = - g = 9.81 m/s2

∴ − y = ( - u sin α) – 𝟏

𝟐 𝐠 𝐭𝟐

y = (u sin α) + 𝟏

𝟐 𝐠 𝐭𝟐---------- (1)

d) Considering horizontal motion of projectile



x = u cos α x t --------- (2)

Velocity with which it strikes the ground (VB):

VB = u = initial velocity

VH = u cos α

VV = u sin α

We have,

𝐯 = 𝐮 + 𝐚𝐭

-VV = - u sin α – gt

VV = u sin α + gt

VB = √ 𝑽𝑯 𝟐 + 𝑽𝑽

𝟐--------------- (velocity of hit)

Direction of hit:

θ = tan-1 ( 𝑽𝑽

𝑽𝑯 ) ------------ (with the horizontal)

Problems:

1) A stone is thrown from the top of the tower 30 m high at an angle of 400 with horizontal

with velocity 40 kmph. Find when and where the stone will hit a horizontal ground. Also

find the velocity and direction of striking.

Ans: t = 1.85 sec, x = 15.74 m, Vb = 26.67 m/s, θ = 71.390

Chapter no. 7 projectile

Engineering

Transcript of Chapter no. 7 projectile