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
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8.8 m
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19.6 m
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34.6 m
|
57
|
78.3 m
|
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45°
|
2.6 m
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10 m
|
23 m
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40 m
|
62.4 m
|
88
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60°
|
2.2 m
|
8.8 m
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19.5 m
|
34.8
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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
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? :- 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 . 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 ?
90° b. 50° c. 40° d . 45°
8.0 What is the most suitable angle to throw the ball. to get the maximum height. ? 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? 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.
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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.
| 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.
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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.
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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 ?
| 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. |
