Lesson Notes By Weeks and Term - Senior Secondary School 2

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 

1. a horizontal constant velocity 
2. 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

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

iii.    A stone released from a catapult.

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

TERMS ASSOCIATED WITH PROJECTILE

1. 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.
2. The maximum height (H):is defined as the highest vertical distance reached and is measured from the horizontal projection plane.
3. 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

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 p 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

2. 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

2. A stone is shot out from a catapult with an initial velocity of 30m  at an elevation of 60,find
3. the time of flight
4. 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

3. 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=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

5. 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

1. 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

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

WEEKEND ASSIGNMENT

1. 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.
2. 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           

3. 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
4. The range of a projectile projected at  θ  to the horizontal with a velocity U is given by
5. a)  U2 sin 2θ   (b) U2 sin2θ (c) 2 U2 sin2 θ (d) 2U2sin2θ       (e)  U2 sin2 θ

g    2g            g                 g        g       

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

THEORY 

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 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.