PHYSICS SSS TWO

PHYSICS SSS TWO

WEEKS ONE & TWO

MOTION UNDER GRAVITY

PROJECTILES: A body that has no motive power of its own will travel under the action of gravity and air resistance when it is launched into space. Such an object is called a projectile or a ballistic missile.
    Hence a projectile can be defined as an object or a body launched into the air or space and allowed to move on its own or move freely under the influence of gravity. Examples are thrown ball or kicked ball, jumping insects, shot arrow, fired bullet and thrown javelin.
The path followed by a projectile is known as its trajectory.

Force of Gravity

This is the pull exerted by the earth on all bodies near it. It is known to be constant at a particular place on the earth’s surface but it varies with observatory height above sea level. Thus, force of gravity varies due to:      (i)shape of the earth  (ii) rotation of the earth.

VERTICAL PROJECTION
    When a body like a ball is thrown vertically upward, the maximum height attained depends on the initial upward velocity.
Given that H=maximum height attained
                    u= initial upward velocity
                     g=acceleration of free fall.
At time (t) after projection, the upward velocity (v) is give by the first equation of motion as:
V = u + at , which will become
V = u-gt (u = -g for upward motion).
At the maximum height, the final velocity V = 0. So  the equation now becomes
0 = u – gt or u = gt
thus, t = u/g   (time to reach the maximum height)
 The time to return to initial point of projection is the same as time to reach greatest height, so total time for the journey known as the time of flight is given by:
T = t + t = 2t, that is 2u/g.
Hence, total time of flight is
T = 2u/g.


Maximum Height Attained:
This is obtained from the second equation of motion as:
S = ut + ½at².  ( here S = H    and  a = -g for upward motion).
H = ut – ½gt² , where t is the time to reach maximum height.
So  H = u x u/g – ½g x u²/g² .
          = u²/g – u²/2g = u²/2g
Thus,  H = u²/2g
OR, from  t = u/g,  u = gt
then  H = (gt)²/2g  =  gt²/2
hence, H = u²/2g  or H = ½,gt²

Example: A body is projected vertically upward with a velocity of 9.8ms^-1. How high does it travel before it comes to rest at the top of its motion? What is the duration of the whole flight? (g = 9.78ms^-2)

Solution:  u = 9.8ms^-1   g = 9.78ms^-2     H = ?

From H = u²/2g

           H = 9.8²/2 x 9.78 = 4.910002045

  Maximum height  H = 4.91m 

Duration of the flight T = 2u/g

                                            = 2 x 9.8/9.78 = 4.00409

 Time of flight T = 2.004secs

HORIZONTAL PROJECTION 

If we roll a ball at a fast speed along a horizontal table, after reaching the edge of the table; it continues its horizontal motion also experiences the pull of gravity. The downward pull will deflect the ball from its original position and finally strike the ground at a distance depending on its original horizontal speed and the height of the table. The motion of the ball is a combination of two motions:

(i)                Horizontal motion at constant velocity.

(ii)              Vertical motion under constant acceleration.

    From the above diagram, the ball is projected from a point A and touches the ground at C.

In the second equation of motion; S = ut + ½at², the distance S is called the range R of the projectile, defined as the horizontal distance of the projectile from the point of projection to where the projectile hits the ground. It could be obtained with acceleration a = g = 0 as follows:

                  R = ut + ½gt²     (g = 0)

                  R = ut (t is the time of flight denoted by T)

Hence       R = uT

Also, the height of projection (i.e the height of the table in this case) can be obtained from the second equation of motion

         S = H = ut + 1/2gt² (u = 0 for downward motion and t is the time of flight  T)

               H = ½gT²

Time of flight T:

From H = ½gT²

          gT² = 2H

   then T =√(2H/g)

   Since R = uT and T = √(2H/g),

   then R = uT or R = u√(2H/g)

Example: A boy throws a stone horizontally with a velocity of 20ms^-1 from the top of a building 40m high. How far from the foot of the building will the stone strike the ground (g = 9.8ms^-2)

Solution:

U =20ms^-1   H = 40m   g = 9.8ms^-2   R = ?

R = uT  = u√(2H/g)

             = 20 x √(2 x 40/9.8) = 57.143

R = 57.14m

ASSIGNMENT
(1)  A ball is released from a height above the ground, find its velocity after `15secs given that g = 10ms^-2
(2) A tennis ball is projected horizontally from the top of a cliff 50m high with a velocity of 10ms^-1 , find
(a) the time taken to reach the ground
(b) the distance from the foot of the cliff to where the ball hits the ground.

3. A student throws a stone horizontally from the top of a building 45m high with a velocity 10ms^-1, determine

(a) the time it will take the ball to strike the ground

(b) the distance of the from the base of the building  when it hits the ground.