**SUBJECT: PHYSICS**

**CLASS: SS 2**

**DATE:**

**TERM: 1st TERM**

**REFERENCE TEXT**

- New School Physics by M.W Anyakoha
- SSCE WAEC Past Questions
- UTME Past Questions

WEEK FOUR

**TOPIC: PROJECTILES AND FALLING BODIES**

**CONTENT**

- Terms associated with projectiles
- Equation of projectile motion
- Uses of projectile

A** PROJECTILE** is an object or body launched into the air and allowed to move on its own or move freely under gravity.

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

Projectile is a two dimensional motion of an object thrown obliquely into the air, the part followed by a projectile is called a trajectory

The following are examples of projectile motion

- A throw rubber ball rebouncing from a wall.
- An athlete doing the high jump.

iii. A stone released from a catapult.

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

**TERMS ASSOCIATED WITH PROJECTILE**

- Time of flight (T):The time of flight of a projectile is the time required for the projectile to get to maximum height and return to the same level from which it projected.
- The maximum height (H):is defined as the highest vertical distance reached and is measured from the horizontal projection plane.
- 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.

Uy = U sin θ (vertical component)

Ux = U cos θ (Horizontal component)

t = time to reach the greatest height (s)

V = u + at v =o, a = -g

θ= u sin – gt

t = U sin θ………………………………………….. 1

g

T = Time of flight (s)

**DETERMINATION OF TIME OF FLIGHT (T), RANGE (R) AND MAXTMUM HEIGHT**

Assuming that Q is the point where the particles meet the target. Let T be the time of flight at Q, the vertical displacement is zero

Vertically S=U sin θ t - 1/2 gt2

0 = U sin θ – ½ gt2

½ gt2 = U sin θ t

T = 2 U sin θ ....................... 2

g

Horizontally, considering the range covered

R= U2 sin 2θ ………………………………. 3

g

For max .range θ = 450

Sin2θ = sin 2 (45) = sin 900 = 1

R= U2

g

R max = U2 ……………………………………………. 4

g

For maximum height H ,

V2 = U2sin2θ - 2g H

At max height H, V=0

H = U2 sin2θ …………………………………………… 5

2g

**APPLICATION 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 p 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= 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

wheret 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

= 1000 x10√6 = 2.449x104 m

buttanθ = s = 2.449 x 104

3,000 x 3,000

θ =83.020

- A stone is shot out from a catapult with an initial velocity of 30m at an elevation of 60,find
- the time of flight
- the maximum height attained

c the range

T = 2U sin θ

g

T= 2 x 30 sin 600

10

T= 5.2s

The maximum height,

H=U2 sin2 θ

H = 302 sin2 (60) = 33.75 m

20

The range ,R =U2 sin 2θ

g

R = 302 sin 2 (60)

10

= 90 sin 120

= 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+1/2gt2

a=g, u=o

120= ½ (10)t2

t2 = 24

t = 24

t =4.9s

(b) Range =u cosθ x T.

but in this case θ =o

cos o =1

R =ut

= 60x 4.9

=294m

s =ut+1/2gt2

a=g, u=o

120= ½ (10)t2

t2 = 24

t = 24

t =4.9s

4 A stone is projected horizontally with a speed of 10m/s from the top of a tower. With what speed does the stone strike the ground?

T = √ 2H/g =√2x50/10 =10

R = ut = 10√10 m

V2=u2 + 2gh

=02+2x10x50 (U= 0 when it strikes the ground)

=0+1000

V =1000

V =33.33m/s

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

i the time of flight

- the maximum height attained and the time taken to reach the height

iii. the velocity of projection 2 seconds after being fired (g = 10m/s)

O=60

u =80m/s

T =? H =? T =? R=?

T = 2 U sin θ

g

T = 2x80 sin 60 = 13 .86 s

10

H = u2 sin 2θ

2g

H = 80 x 80 x sin60 =240 m

20

t = U sin θ = 80 sin 60 = 6.93 s

g 10

R =U2 sin 2 θ = 802sin2 (60)

g 10

R = 640 sin 120 = 554.3m

(iii) Vy = U sin θ – gt

Vy = 80 sin 60 – 20 = 49.28m/s

Ux = U cos θ

Ux = 80 cos60 = 40 m/s

U2 = U2y + U2x

= 49.282 + 402 = √1600+ 2420 = 63.41 m/s

**GENERAL EVALUATION**

A stone of mass 0.4Kg is attached to a string of length 2.5m and its is spin around by a boy at 5rad/s. calculate

- The force necessary for this motion.
- The linear velocity with the stone moves.

**WEEKEND ASSIGNMENT**

- 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) 200m (b) 65m (c) 250m (d) 100m.
- 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 sin θ (c) U2 sin θ (d) 2U2sin2θ (e) 2U sin2 θ

g 2g g g g

- 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
- The range of a projectile projected at θ to the horizontal with a velocity U is given by
- a) U2 sin 2θ (b) U2 sin2θ (c) 2 U2 sin2 θ (d) 2U2sin2θ (e) U2 sin2 θ

g 2g g g g

- For a projectile the maximum range is obtained when the angle of projection is;
- a) 60 b) 30 c) 45 d) 75 e) 90

**THEORY **

- 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?
- 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.

**READING ASSIGNMENT**

New Sch. Physics for Senior Sec. Schls. Pages 137-144.

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