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Projectiles.

posted Feb 10, 2015, 9:59 PM by Ranmini Perera   [ updated Jul 6, 2018, 4:05 PM by Upali Salpadoru ]
Motion -Projectiles

      If you drop or throw an object or eject in any other way you would have launched a projectile.
If a force is applied to a projectile, it would be only at the beginning. Thereafter it would continue to move due to inertia (assuming no drag)  guided only by the force of gravity. A stone from a catapult, a bullet from a gun, a rocket with the motors shut  or an athlete airborne for long jump or high jump  are a few examples of projectiles. 

Fig.1. Kumar Sangakkara projecting the ball to the ropes.

     If you know the initial velocity of the object , as the acceleration due to gravity is uniform it is possible to determine the time it would remain in air, maximum height it would reach and the horizontal displacement or even to plot the complete trajectory.  Of course our calculations will be slightly off the mark as we neglect the air resistance which may be appreciable for certain objects.
Let us consider a few examples.


Example 1.0   

A Prince hurls a climbing rope vertically up to the King’s daughter in the solitary tower. If the velocity of the throw is  25ms-1 will it reach the princess 28 m.  above the ground? Use the acceleration due to gravity as 10 m-s2    and a drag equivalent to  0.5 ms2 .

WORKING

      Acceleration   =  change in velocity/ time

                       a.= ( V2 – V1) / t

           -10 - 0.5  =  (0 -  V1) / t

                        t.= 25 /- 10.5    = 2.4  seconds

Height =  average velocity x time

        h. =  25/2  x  2.4      = 30 m

Yes. The Princess can get the rope.

Example 2.0

Fig. Kumar Sangakkara projecting the ball to the ropes.


Kumar Sangakkara slammed the ball attempting a six.  The distance to the boundary line was 100 m.  The ball left at 34 ms-1 at an angle of  45° would the ball clear the line?


The vertical component of speed  

      =   34 x sin 45° 
      = 34 x 0.7071   =   24.14 ms-1

Using formula:-                    

acceleration (g) =  Change in velocity/ time
                                                          -10   = (V2 –V1) / t        

                                                              t  x -10 =  (V2 –V1)
                                                             t.   =  ( 0 – 24.14)  / -10 
                                                            t     =  2.41 s.


 The horizontal component of speed =  34 cos 45°=  0x 0.7071  

                                                         = 24.1 ms-1
Assuming the velocity does not change
                                                    d =  V/ t
                                                    d = 24.1x 2.41 = 58 m
This is the first half of the parabola.  The distance traveled as the ball is rising.  Total distance is double this.
The distance for the complete parabola = 2 x 58  =  116 m. 
                                                                             So Cheers !  he gets a sixer !

2.0 

George using a sling shot (catapult) hurled a stone at a  speed of 40 ms-1 and an angle of 30° to the ground.  Calculate the height and the horizontal distance it should travel, neglecting air resistance.



The velocity vector may be resolved into horizontal and vertical components.  The values are shown in the figure above.


As sine 30° is  0.5  V  becomes  20 ms-1.
Using the formula :-

acceleration =   change in velocity /  time
    -10 m s-2 =  (V2 – V 1 )  / t
   -10 m s-2  =   (0 – 20 ms-1)  / t
                                                      t  =    - 20  ms-1 / - 10  s-2
                                                      t  =   2 s


From this we can find the maximum height it will reach:-
                              height.  = average velocity x time
                                     h.    = ½ x20 ms-1 x t 
                                     h     =  10 ms-1  x 2  s   
                                     h     =  20 m


The time for the stone to go up and come down  2x2  = 4  s.
The horizontal distance during that time    d=  v x t
                                                                              =40 cos 30°  ms-1   x  4s
                                                                              = 40 x 0.866 x4 m       
                                                                              = 138.56 m   

 

 Example 3.0


 
 

The teenage Olivia Mc Traggart of New Zealand clears the bar at 4.2 m in paulvault.

  1. Find the time to go up to the bar.

     d. = V‍f .t  + ½ . - a . t2

            d, = 0 + - (-10/2) x t2

               ∴    t2  =  d X 5

                    2 =  4.2 / 5..= 0.84

                               t=0.9 s.

2. Find the minimum vertical,component of the  initial velocity required to clear the bar.

 d= (Vf + Vi) / 2 x t.

d = V a Xt

ie: Va = d/t

= 4.2 / 0.9.......= 4.7 ms-1

3. If she takes off at an angle of 80° find the initial velocity in that direction.

     If R is the required velocity ,

     Vertical velocity = R sin 80°.

     ∴ R sine 80 = 4.7

               R = 4.7 / 0.98

               R= 4.8 ms-1


Here are the distances theoretically possible for throwing objects such as shot put, discus and javelin.

     

 The graph showing throwing velocity  and the possible displacement.

  Thrown at an initial velocity of 20 degrees.

Throwing angle and the height of the projectile when thrown at a velocity of 20ms-1.

  The values for the above graph was from this table.



Angle to horizontal

Vertically Resolving

Finding the Height.


TIME

Horizontal resolution.

Finding the displacement.

Velocity initial

20 sinΘ

ms-1


Average velocity

Va=V1+V2

        2

ms-1

Height

Va x t




Meters

Time.

t = V /g




Seconds

Velocity

20 cos Θ

ms-1

Displacement


d= v.2 t


Meters

10°

20x0.17

=  3.5  

1.7

0.6  m

0.35x 2

= 0.70

20 x 0.98

=  19.6

16.8 x 0.34  2 m.

13. 7 m.

20°

20 x 0.34

= 6.84

3.4

2.3  m

0.68 x2

= 1.36

20x 0.94

=18.8  

18.8 x 0.68 x2

25.6  .m.

30°

20x0.5   

=10

5.0

5.0 m

1.00 x2

= 2.0

20 x 0.87

17.4

17.4 x 1

34.8m.

40°

20x0.64

= 12.8

6.4

7.2 m

1.28 x2

=  2.56

20 x 0.77

15.4

15.4 x 1.28 x2

39.4 m.

45

20 x 0.71

  14.2

7.1

10.1 m

1.42 x 2

=  2.84

20 x 0.71

14.1

14.1 x 1.42 x2

40.2

50°

20x0.77

= 15.4

7.7

11.8 m

1.54 X2

=3.08

20 x 0.64

12.9

12.9 x 1.54

39,8  m

60°

20x0.87

= 17.4

8.7

14.8 m

1.74

20 x 0.5

10

10 x 1.74 x2

34.8 m

70°

20x0.94

= 18.8

9.4

17.6 m

1.88

20 x 0.34

6.8

6.8  x1.88x2

25.5  m

80°

20x0.98

= 19.6

9.8

19.2 m

1.96

20 x 0.17

3.5

3.5 x 1.96 x 2

13.6  m

90°

20x1.00

= 20

10

20 m


2.00

20 x 0

0.

0. 0  m.


 

Angle

5 ms-1

10 ms-1

15 ms-1

20 ms-1

25 ms-1

30 ms-1

30°

2.6 m

8.8 m

19.6 m

34.6 m

57

78.3 m

45°

2.6 m

10 m

23 m

40 m

62.4 m

88

60°

2.2 m

8.8 m

19.5 m

34.8

52.6 m

69 m



Throwing angle and the height of the projectile when thrown at a velocity of 20ms-1.


    The readings used for the above graph

 


 Multiple Choice

      Kinematic equations.

a.  d= Va xt.              b.Va =  Vi + Vf  / t.

  c.Vf  = Vi + axt.       d..Vf2 = Vi 2 + 2 a.d.

e. d- Vi xt + 1/2 at2

  • Use ‘g’ as 10 ms-2

  • Throwing angle is the angle with the horizontal.

  • Neglect the drag unless asked.


   Question Answer
 

     

1.0   If you throw a ball up at a velocity of 22.5 m/s, what will be the velocity at the maximum height? 

A:-0 m/s    b. :-2.25m/s  c:- 22.5, d:- 11.25.m/s.

2.0 What will be its velocity when it  reaches the ground?

  1. :- 22.5 m/s.  B:-Higher than 22.5m/s.  C:- ;Lower than 22.5m/s. D:- 32.5 m/s


 Data for Q. 3 to 5.

Hema throws a ball   30° to the horizontal at  

Vi =10m/s.  

3.0 What will be the initial velocity upwards?

      A:- 10m/s.  B:-10. Cos C:- 10.Sine 30°.D:- 10 .Sin 60°,,      

4.0 What will be the horizontal velocity?

    A:- 10m/s. B:-10. Cos30°, C:-  10.Sine 30. D:- 10 .Sin 60°,

5.0 What will be the final horizontal velocity as it touches the ground?

     a.10m/s b.  0 c. 10 sine 30°    d. 10. Cos30°

6.0 A ball of 2 kg mass is dropped from a height and it .

      falls for 12 seconds what will be the final velocity .

  1. 240 ms-1.  B. 120 ms-1.  C. 240 ms-1    d.  30 ms-

7.0 What is the most suitable angle to the horizontal, to throw a ball to get the maximum distance ?
  1. 90°   b. 50°     c. 40° d . 45°

8.0
 What is the most suitable angle to throw the ball.

     to get the maximum height. ?

  1. 90°   b. 50°     c. 40° d . 45°

 9.0 If Jane threw a ball weighing 500N  vertically up , at a velocity of 20ms-1,  how high will it go?

  1. 10 m.   b.20. m c. 30, m. d. 40. m.

10.0She catches the ball as it comes down.

      What would have been its velocity.

       a.10ms-1. b.-20ms -1.c. -30ms-1. D.40ms-1.

Highlight to get Answers.

 
  1. a


  1. b.

(Consider your height)




  1. c.


  1. b.



  1. d


  1. b.


  1. d


  1. a.



  1. b



  1. b


Question  1.0

Nirosha fires a gun at an angle of 30°. The rubber bullet starts with a velocity of 20ms-1 .  


1.1 Find the vertical and horizontal components of the velocity.

1.2.Find the total time it will remain in air.

1.3   Find the average upward velocity

1.4  Find the height and horizontal displacement.




Marks:- 4 x 4 =16



Q. 1.0

1.1     

Vertical =20 sin30

= 20 x 0.5 ………..= 10 m/s.

Horizontal

= 20 cos 30

= 20 x 0.8660………..=  17.32  m/s

1.2      

time to climb

                        = V /a

                      t = 10/ 10

                          Total time = 2 s.

1.3       

           Va = (Vf + Vi)/2.

               =(0 + 10)/2……….. =5 m/s.

1.4  

Height = Va xt.

     H = 5 x2 …………...= 10 m.

        

     D = V x t

       = 19.32x 2…………. =      38.64 m.

      


Question  2.0

There is a ripe mango in a tree at a height of 12 meters.  If Jane throws an object at a velocity of 20ms-1 using an angle of 60°  Find the following.

2.1  Vertical component of velocity.

2.2  Time to climb.

2.3  Maximum height.

2.4  While Jane selected a stone as the missile, her brother John used a stick. Give one advantage of each weapon.

2.5 will it be possible to get the fruit?

2.0

2.1

Vertical component

   = 20 sin 60

   =20 .8660 …………..=17.32  m/s

2.2 Time to climb

   = 17.32/ 10 ………..= 1.732 s.

2.3 Height =Va x t

    8.66 x 1.73 …………=.14.7 m

2.4  stone :- low air resistance.Stick: Better chance of hitting the fruit.

2.5.  Yes.


Marks :- 2x5 = 10.



Question 3.0



When Raman  throws a cricket ball at an angle of 45° , he gets an average height of 5 meters.  Find the following:-

   3.1 Initial vertical velocity.

   3.2 Climbing time.

   3.3 Throwing velocity at 45°.

   3.4 Average vertical velocity.

   3.5 Horzontal displacement.


What is the initial velocity of the throw?

3.1  Using  Kinetic Equation d.

Vertical velocity.

           Vf2 = Vi2 + 2 a d.

           0 = Vf2+2 x-10.5

           Vi2= 100

                         Vi = 10 ms-1

3.2 Climbing time.

            t = V / a

             t= 10/10…. 1sec.

3.3 Velocity at 45°

            R sin 45° = 10 m/s

             R x 0.7071= 10

             R= 14,1m/s

3.4   Vav. = 10/2

            = 5 m/s.

3.5 Displacement

     = Va x total time.

     = 5 x 2…………….= 10 m.

     

Marks:-  3x5 = 15.



Questioin 4.0

An aeroplane flying at 150 ms-1 at an altitude of 2000 m. had to cut off the engines completely.

Find the following:-

4.1 Minimum time available to be in air neglecting lift and drag.

4.2 If you consider the lift and drag will there be more time?

4.3 What is the ground cover it can get neglecting lift and drag ?


Marks:- 4 x3 =12


4.1

Kinematic equation.. e.

  d= Vi xt + 1/2 at2

    2000 = 0 + 0.5 x10 x t2

       T2  = 2000/ 5

    t= 20 seconds.

4.2  As lift and drag will oppose the falling acceleration there will be  more time.

4.3    d =Vxt

        d = 150 x 20

         =3000 m.


 

Question 5.




Valerie Adams is a shot put World champion. If she throws the put at a velocity of 14 ms-1 at an angle of 40° to the horizontal  Find the following:-


5.1  Vertical and Horizontal components.

5.2  Time the object will be in air.

5.3  Maximum height it will reach.

5.4 The displacement she will get.

5.5 If she throws from a height of 2 meters above the ground

    What will be the velocity when it comes down? (HELP- Use kinematic equation ………)

5.6 What is the additional time the putt will take?

5.7 What is the extra displacement she will get?


 

m/s.

Vf2 = Vi2+ 2.a.d.

Vf2   = 92  + 2x10x2

Vf2   = 81 + 40

Vf  =  √121…..=11ms-1

5.6.   

   Va=   (11 +9)/2  =10 m/s

   Va =  d/t

   10 = 2/t

      t= 0.2 seconds.

5.7  Displacement = Vh x t

                          =10.7 x0.2

                           2.14 m.













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