**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

- a horizontal constant velocity
- a vertical free fall due to gravity

Examples of projectile motion are the motion of;

- a thrown rubber ball re-bouncing from a wall
- An athlete doing the high jump

- A stone released from a catapult

- A bullet fired from a gum
- A cricket ball thrown against a vertical wall.

U y ** **

H_{max}

t t

)θ

P U_{x }Q

U_{y} = U sin θ (vertical component) ——————- 1

U_{x} = U cos θ (horizontal component) ——————- 2

## TERMS ASSOCIATED WITH PROJECTILE

*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 = 2__U sin θ__ ——————- 4

g

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

For maximum height H,

V^{2} = U^{2} sin^{2}θ – 2g H

At maximum height H, V=0

2gH = __U ^{2} __ ——————- 5

2g

*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 + ½at^{2 }(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 = __2U ^{2} sin θ cos θ__

g

From Trigonometry function

2 sin θ cos θ = sin 2θ

R= __U ^{2} sin 2θ__

g

For maximum range θ = 45^{0}

Sin2θ = sin 2 (45) = sin 90^{0} = 1

R= __U ^{2}__

g

R_{max} = __U ^{2}__

g

## USE OF PROJECTILES

- To launch missiles in modern warfare
- 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

- 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= 10^{3 }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 gt^{2 }(u=o)

h =1/2gt^{2}

Where: t is the time the bomb takes to reach the ground: 300=1/2gt^{2}

t^{2}= 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 = 2.449×10^{4} m

But tanθ = __ s__ = __2.449 x 10 ^{4}__

3,000 3,000

θ =83.02^{0}

- A stone is shot out from a catapult with an initial velocity of 30m at an elevation of 60
^{0}. Find - the time of flight
- the maximum height attained
- the range
- The time of flight

T = __2U sin θ__

g

T= __2 x 30 sin 60 ^{0}__

10

T= 5.2s

- The maximum height,

H=__U ^{2} sin^{2} θ__

2g

H = __30 ^{2} sin^{2} (60)__

20

__H __= __33.75 m__

- The range,

R = __U ^{2}sin 2θ__

g

R = __30 ^{2} sin ^{2} (60)__

10

R = 90 sin 120

R = __77.9 m__

- 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

- s =ut+1/2gt
^{2}

a=g, u=0

120= ½ (10)t^{2}

t^{2} = 24

t = 24

t = 4.9s

- Range =u cosθ x T.

But in this case θ = 0

Cos 0 =1

R = ut

R = 60x 4.9

R =294m

- 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**

v^{2}=u^{2} + 2gh

v^{2}=10^{2}+ (2x10x50)

v^{2}=100+1000

v^{2}=1100

v^{2}=33.17m/s

- A projectile is fired at an angle of 60 with the horizontal with an initial velocity of 80m/s. Calculate:
- the time of flight
- 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

- T =
__2 U sin θ__

g

T = __2×80 sin 60__

10

T = 13 .86 s

- H =
__u__^{2}sin 2θ

2g

H = __80 x 80 x sin60__

20

H = 240 m

- t =
__U sin θ__

g

t = __80 sin 60__

10

t = 6.93 s

R =U^{2} sin 2 θ

g

R = 80^{2}sin2 (60)

10

R = 640 sin 120

R = 554.3m

- V
_{y}= U sin θ – gt

V_{y} = 80 sin 60 – 20

V_{y} = __49.28m/s__

U_{x} = U cos θ

U_{x} = 80 cos 60

U_{x} = 40 m/s

U^{2} = U^{2}y + U^{2}x

U^{2} = 49.28^{2} + 40^{2 }

U = √ (1600+ 2420)

U = 63.41 m/s

## ONLINE WORK

- (a) Define the term projectile (b) mention two application of projectiles
- 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
- 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**

- 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
- 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
- For a projectile the maximum range is obtained when the angle of projection is; (a) 60
^{0}(b) 30^{0}(c) 45^{0}(d) 90^{0} - 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 sin ^{2} θ__ (b)

__U__(c)

^{2}sin^{2}θ__U__(d)

^{2}sin θ__2U__

^{2}sin^{2}θg 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^{0} to the level ground. Calculate the:

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

**SECTION B **

- A gun fires a shell at an angle of elevation of 30
^{0}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? - (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
^{0}above the ground level. Calculate the maximum height attained by the object

See also

DERIVATION OF EQUATONS OF LINEAR MOTION