# PROJECTILES AND ITS APPLICATION

## MEANING OF PROJECTILE

A projectile motion is one that follows a curved or parabolic path .It is due to two independent motions at right angle to each other .These motions are

1. a horizontal constant velocity
2. a vertical free fall due  to gravity

Examples of projectile motion are the motion of;

1. a thrown rubber ball re-bouncing from a wall
2. An athlete doing the high jump
• A stone released from a catapult
1. A bullet fired from a gum
2. A cricket ball thrown against a vertical wall.

U y

Hmax

t                                     t

P                                    Ux                                         Q

Uy = U sin θ                   (vertical component)      ——————- 1

Ux = U cos θ         (horizontal component) ——————- 2

## TERMS ASSOCIATED WITH PROJECTILE

1. Time of flight (T) – The time of flight of a projectile is the time required for it to return to the same level from which it projected.

t= time to reach the greatest height

V = u + at   (but, v =o, a = -g)

θ= u sin – gt

t = U sin θ                      ——————- 3

g

T = 2t = 2U sin θ           ——————- 4

g

1. The maximum height (H) – is defined as the highest vertical distance reached measured from the horizontal projection plane.

For maximum height H,

V2 = U2 sin2θ – 2g H

At maximum height H, V=0

2gH = U2                       ——————- 5

2g

1. The range (R) – is the horizontal distance from the point of projection of a particle to the point where the particle hit the projection plane again.

Horizontally, considering the range covered

Using S= ut + ½at2 (where a=0 for the horizontal motion)

OR

S = R = U cosθ x t (distance = velocity x time; there time is the time of flight)

R = U cos θ (2 U sin θ)

g

R = 2U2 sin θ cos θ

g

From Trigonometry function

2 sin θ cos θ = sin 2θ

R= U2 sin 2θ

g

For maximum range θ = 45

Sin2θ = sin 2 (45) = sin 90 = 1

R= U2

g

Rmax = U2

g

## USE OF PROJECTILES

1. To launch missiles in modern warfare
2. To give athletes maximum takeoff speed at meets

In artillery warfare, in order to strike a specified target, the bomb must be released when the target appears at the angle of depression φ given by:

Tan φ =1/u  √gh/2

EXAMPLES

1. A bomber on a military mission is flying horizontally at a height of zoom above the ground at 60kmmin-1. lt drops a bomb on a target on the ground. Determine the acute angle between the vertical and the line joining the bomber and the tangent at the instant. The bomb is released

Ux                  60m/ min

3,000m

Horizontal velocity of bomber = 60km/min= 103 ms-1

Bomb falls with a vertical acceleration of g = 10m/s

At the release of the bomb, it moves with a horizontal velocity equals that of the aircraft i.e. 1000m/s

Considering the vertical motion of the bomb we have

h =ut+1/2 gt2 (u=o)

h =1/2gt2

Where: t is the time the bomb takes to reach the ground: 300=1/2gt2

t2= 600

t=10√6 sec

Considering the horizontal motion we have that horizontal distance moved by the bomb in time t is given by

s =horizontal velocity x time

s = 1000 x10√6

s = 2.449×104 m

But tanθ =  s   = 2.449 x 104

3,000      3,000

θ =83.02

1. A stone is shot out from a catapult with an initial velocity of 30m at an elevation of 60. Find
2. the time of flight
3. the maximum height attained
4. the range
5. The time of flight

T = 2U sin θ

g

T= 2 x 30 sin 60

10

T= 5.2s

1. The maximum height,

H=U2 sin2 θ

2g

H = 302 sin2 (60)

20

H = 33.75 m

1. The range,

R = U2sin 2θ

g

R = 302 sin 2 (60)

10

R = 90 sin 120

R = 77.9 m

1. A body is projected horizontally with a velocity of 60m/s from the top of a mast 120m above the grand, calculate

(i) Time of flight, and (ii) Range

60 m/s

120

R

1. s =ut+1/2gt2

a=g, u=0

120= ½ (10)t2

t2 = 24

t = 24

t = 4.9s

1. Range =u cosθ x T.

But in this case θ = 0

Cos 0 =1

R = ut

R = 60x 4.9

R =294m

1. A stone is projected horizontally with a speed of 10m/s from the top of a tower 50m high and with what speed does the stone strike the ground?

Solution

v2=u2 + 2gh

v2=102+ (2x10x50)

v2=100+1000

v2=1100

v2=33.17m/s

1. A projectile is fired at an angle of 60 with the horizontal with an initial velocity of 80m/s. Calculate:
2. the time of flight
3. the maximum height attained and the time taken to reach the height
• the velocity of projection 2 seconds after being fired (g = 10m/s)

θ =60; u =80m/s

1. T = 2 U sin θ

g

T = 2×80 sin 60

10

T = 13 .86 s

1. H = u2 sin 2θ

2g

H = 80 x 80 x sin60

20

H = 240 m

1. t = U sin θ

g

t = 80 sin 60

10

t =   6.93 s

R =U2 sin 2 θ

g

R = 802sin2 (60)

10

R = 640 sin 120

R = 554.3m

• Vy = U sin θ – gt

Vy = 80 sin 60 – 20

Vy = 49.28m/s

Ux = U cos θ

Ux = 80 cos 60

Ux = 40 m/s

U2 = U2y + U2x

U2 = 49.282 + 402

U = √ (1600+ 2420)

U = 63.41 m/s

## ONLINE WORK

1. (a) Define the term projectile (b) mention two application of projectiles
2. A ball is projected horizontally from the top of a hill with a velocity of 30m/s. if it reaches the ground 5 seconds later, the height of the hill is
3. A stone propelled from a catapult with a speed of 50m/s attains a height of 100m. Calculate: a. the time of flight b. the angle of projection c. the range attained.

## ASSIGNMENT

SECTION A

1. A stone is projected at an angle 60 and an initial velocity of 20m/s determine the time of flight (a) 34.6s (b) 3.46s (c) 1.73s (d) 17.3s (e) 6.92s
2. A stone is projected at an angle 60 and an initial velocity of 20m/s determine the time of flight (a) 34.6s (b) 3.46s (c) 1.73s (d) 17.3s (e) 6.92s
3. For a projectile the maximum range is obtained when the angle of projection is; (a) 60 (b) 30 (c) 45 (d) 90
4. The maximum height of a projectile projected with an angle of to the horizontal and an initial velocity of U is given by

(a)  U sin2 θ    (b) U2 sin2θ (c) U2sin θ (d) 2U2sin2θ

g                 2g               g                 g

Use this information to answer questions 5 and 6: An arrow is shot into space with a speed of 125m/s at an angle of 15 to the level ground. Calculate the:

1. Time of flight (a) 5seconds (b) 6.47seconds (c) 16.01seconds (d) 4.7seconds
2. Range of the arrow (a) 350m (b) 781.25m (c) 900m (d) 250.71

SECTION B

1. A gun fires a shell at an angle of elevation of 30 with a velocity of 2x10m. What are the horizontal and vertical components of the velocity? What is the range of the shell? How high will it rise?
2. (a) What is meant by the range of a projectile? (b) An object is projected into the air with a speed of 50m/s at an angle of 30 above the ground level. Calculate the maximum height attained by the object