PHYSICS S.S ONE 3RD TERM WEEK 8

 

Topic:                    PROJECTILES     

Sub topic:

Reference materials:

(1) ESSENTIAL PHYSICS, TONALD PUBLISHERS, EWELUKWA (2017)

(2) NEW SCHOOL PHYSICS, AFRICAN FIRST PUBLISHERS, ANYAKOHA M.W (2011)

(3) COMPREHENSIVE PHYSICS FOR SSS, EZEBUIRO G.N (2004)

(3) INTERNET

Instructional materials: Rubber ball, catapult, small pebbles

Entry behavior: The students have been taught expansion

Behavioural objective: At the end of the lesson the students should be able to:

I.                    Identify a projectile motion

II.                  Derive the time of flight, maximum height and range of a projectile

III.                Solve simple problems on projectiles

IV.                State the application of projectile in sport and warfare.

CONTENT

PROJECTILES

An object that moves through space on projection is called a projectile. The path taken is called trajectory.

Example of projectile motion are the motion of:

1.       A thrown rubber ball bouncing from a wall

2.       An athlete doing the high jump.

3.       A stone released from a catapult

4.       A bullet fired from a gun

MOTION OF A PROJECTILE

I.                    VERTICAL PROJECTION

If a stone or ball is projected vertically upwards, its projection velocity to decrease more slowly until it comes momentarily to a stop. It then falls back faster and faster till it hits the plane of projection.

 

 

 

 

               

 

 

 

 

 

The decrease in velocity as the ball rises higher up, is due to the effect of gravity. However, the maximum height on object under this condition can attain depends on the magnitude of the initial upward velocity used in projecting it.

 If h_max = max. height attained, u = initial upward velocity, g = acceleration of free fall.

Then, the Time ,t, taken to reach the max. height would be obtained from V = U – gt.  At h_{max ,}V = 0

                0 = U – gt

                U = gt,                   t = U/g

To determine the time of flight (the time required from  the object to be in air, this time, T, is twice the time required for the object to reach the h_max.

                T = 2t

                T = 2U/g

To determine the h_max attained used V² + U² -2g h_max

At h_max, V = 0

                0 = U² – 2g h_max

                U² = 2g h_max

                h_max = U²/2g.

II.                  HORIZONTAL PROJECTION

 

 

 

 

 

 

 

When an object is projected horizontally with a horizontal velocity U, from a height, h, it will:

i.                     Describe a path that is half parabola as shown above.

ii.                   It takes the same time to reach the ground as though, it is dropped vertically from the same height.

To determine the height, h, from which an object is projected.

By applying the equation h = Ut + 1/2gt² and note that “U” along h is 0.

                                                h = 1/2gt²

The time to reach the ground would be t =

To determine the horizontal distance (Range) covered with the horizontal velocity, U; g in the horizontal direction

½ gt² = 0, by taking h = Range ( R) , then equation becomes  R = Ut + 0

                                Recall,   t =   , therefore     R =

Resultant velocity of a projectiles

   When a projectile is in flight, its resultant velocity, V, which is tangential to the direction of flight is made up of :

i.                     Horizontal component velocity, Vx = horizontal velocity of projection, u.

ii.                   Vertical component velocity, Vy = vertical downward velocity (gt), since U = 0 in the vertical direction.

U = horizontal velocity of projection

 

 

 

 

 

Horizontal projection of ball indicating velocity after time, t.

The resultant velocity after time t,  V = v²_x + v²_y

                                                                      V = v²_x + (gt)²

If V is inclined at an angle of  θ to V_x, then, tan θ =V_y/V_x = gt/u

                                                                                θ = tan^-1 (gt/u)

III.                MOTION DOWN AN INCLINED PLANE

If the friction between an object and the inclined plane in which it is in contact with is neglected, the object would tend to slip down the inclined surface faster because of the component effect of acceleration due to gravity.

 

 

 

 

 

 

By resolution, component of g along the plane g_x = g sin θ  and that perpendicular to the plane g_y = g Cos θ. If S = length of the inclined plane, t = time taken to ship down the plane, θ= angle of inclination of plane, V = final velocity attained on reading the base of plane and noting that U = 0 at rest then final velocity V = g_xt when U= 0

                i.e  V = gt sin θ

                distance covered = S = 1/2g_xt²

                                S = 1/2gt²Sin θ

If  V2 = U2 +2gs is applied, then S = V²/2g_x, since U = 0

                                                                S = V²/2gSin θ

To determine the vertical height of fall: Sin θ =h/s.      h = Sin θ S

Substituting for ” S” using equation         S = V²/2gSin θ

The vertical height of fall, h , becomes  h = V²/2gSin θ X Sin θ

                                               

                                                 h= V²/2g

IV.                INCLINED PROJECTION

 When an object is projected into space at an angle of θ to the horizontal:

i.                     It describes a path that is fully parabolic

ii.                   Its velocity of projection decreases slowly and slowly and it gain height.

iii.                  As the maximum height reached, the velocity becomes zero.

iv.                 Because of gravity, the object returns to same projection plane a distance R from the point of projection.

 

 

 

 

 

 

 

 

The vertical component of velocity Uy = U Sin θ

The horizontal component of velocity Ux = U Cos θ

To determine the time taken to attain the hmax vertical component velocity = Uy

Final velocity at hmax = 0, acceleration of free fall = -g

                From  V = Uy – gt

                                0 = Uy – gt

                                Uy  = gt

                                t  = Uy/g

                                t =

Time of flight, T, = 2t

                         T  = 2

To determine the Hmax

Apply h = Uyt - ²

Initial velocity, U, vertical component velocity required to attain the Hmax = Uy

By applying Hmax = Uyt - ²

And noting that t =  it implies that

Hmax  = U Sin θ – ( ) - g(  )²

            =      -

Hmax =   

To determine the Range, R,: Range is the horizontal distance from the point of projection to the point where the projectile hits the horizontal plane. The velocity that enables the projectile to cover this range is the horizontal component velocity Ux = U Cos θ. Since g = 0 along this direction and time required to cover the range = time of flight, T.

                Then, R = Ux T

                                R =

                            R = 2U²

             Recall, Sin 2 θ = 2Sin θCos θ

                                         R =

                The maximum Range a projectile can attain is when the angle of projection is 45^o.

 

                                     R_max =

                                     R _max=

 

 

 

 

Example

1.       A ball is thrown vertically upward with a velocity of 20m/s. what is the maximum height reached after 30s ?

Solution

U = 20m/s,   g = 10m/s²

                                  H _max=    =       =     = 20m

2.       Felix slides from rest a height of 12.2m down a smooth plane inclined at an angle 30^o to the horizontal. Calculate the distance covered.

Solution

  h =12.2m,  θ = 30^o

 h = Sin θ X S ;  S =      =    =        = 24.4m

3.       A bullet is fired at an angle 45^o to the horizontal with a velocity of 490m/s. what time does it take to reach the highest point ( taking g = 9.8m/s²)

Solution

Θ = 45^o, U = 490m/s, g = 9.8m/s², t = ?

                                     t=   =   =    =   = 35.4s

APPLICATION OF PROJECTILE

1.       Shooting of arrow, guns and rocket.

2.       Sport and games: throwing of javelin and discus, kicking of football and volleyball and throwing of basketball etc

3.       Taking-off and landing of aircraft

4.       Launching of mechanically propelled missiles eg inter-continental ballistic missiles.

PRESENTATION

Step I: The teacher defines projectiles and state some examples of projectile.

Step II: The teacher explains vertical and horizontal projectile.

Step III: The teacher leads the students to derived time of flight, time to reach maximum height, maximum height and range of an inclined projectile.

Step IV: The teacher leads the students to solve problem on projectile

Step V: The teacher lists the application of projectile

EVALUATION

The teacher evaluates the lessons by asking the following questions:

I.                    Define trajectory

II.                  Give two examples of projectile motion

III.                What is the time of flight of a vertical projection

IV.                The maximum angle the maximum range can attain is

V.                  State two application of projectile in warfare

ASSIGNMENT

1.       A stone propelled from a catapult with a speed of 50m/s attain a height of 100m. calculate (a) time of flight (b) angle of projection (c) the range attain. (ans: 846s; 63.43^o; 200m)

2.       A body is projected upward at an angle of 30o with the horizontal at an initial speed of 200m/s. in how many seconds will it reach the ground? How far from the point of projection will it strike? [ans: 20s; 3464m]