Projectile Motion


Published on

This PowerPoint relates to Topic 1 Section 1 of the SACE Course in south Australia

Published in: Education
1 Comment
  • Sir can u email me these slides at
    Are you sure you want to  Yes  No
    Your message goes here
No Downloads
Total views
On SlideShare
From Embeds
Number of Embeds
Embeds 0
No embeds

No notes for slide
  • Again, using the vertical component of the motion a projectile fired upward with a v i V will hit the ground with the a velocity of the same magnitude provided the ground is level, but opposite in direction. Also the time taken to go up is the same as the time for it to come down to the same level again. Thus, we can use the vertical component of the motion of the projectile to find the time to reach maximum height and hence the time of flight.
  • Projectile Motion

    1. 1. Projectile Motion Section 1: Topic 1
    2. 2. Vectors <ul><li>Scalar quantity </li></ul><ul><ul><li>Magnitude only </li></ul></ul><ul><ul><ul><li>Distance </li></ul></ul></ul><ul><ul><ul><li>Energy </li></ul></ul></ul><ul><li>Vector quantity </li></ul><ul><ul><li>Magnitude and direction </li></ul></ul><ul><ul><ul><li>Displacement </li></ul></ul></ul><ul><ul><ul><li>Force </li></ul></ul></ul>
    3. 3. Terms <ul><li>Terms such as speed and velocity : </li></ul><ul><ul><li>are used interchangeably. </li></ul></ul><ul><li>They both have a precise meaning in Physics. </li></ul>
    4. 4. Scalars <ul><li>A car travels at a speed of 80 km h -1 . </li></ul><ul><li>The only information given by the value 80 km h -1 is : </li></ul><ul><ul><li>the magnitude of the speed. </li></ul></ul><ul><li>It does not give any indication of : </li></ul><ul><ul><li>the direction in which the car travelled. </li></ul></ul>
    5. 5. Scalars <ul><li>Speed is an example of a scalar quantity . </li></ul><ul><li>There are many examples of scalar quantities including : </li></ul><ul><ul><li>distance, </li></ul></ul><ul><ul><li>energy and , </li></ul></ul><ul><ul><li>work. </li></ul></ul>
    6. 6. Vectors <ul><li>If a car travels at a velocity of 80 km h -1 North : </li></ul><ul><ul><li>a magnitude and , </li></ul></ul><ul><ul><li>direction are given. </li></ul></ul><ul><li>Velocity is an example of a vector quantity . </li></ul>
    7. 7. Vectors
    8. 8. Vectors <ul><li>Other examples of vector quantities include : </li></ul><ul><ul><li>Displacement , </li></ul></ul><ul><ul><li>Momentum and , </li></ul></ul><ul><ul><li>Force. </li></ul></ul>
    9. 9. Vectors <ul><li>Vector quantities can also show direction by ; </li></ul><ul><ul><li>the use of negative signs; eg ; </li></ul></ul><ul><ul><li>- 80 km h -1 . </li></ul></ul><ul><li>This means the car is travelling at 80 km h -1 ; </li></ul><ul><ul><li>in the opposite direction. </li></ul></ul>
    10. 10. Vectors <ul><ul><li>A directed line segment , </li></ul></ul><ul><ul><li>drawn to scale </li></ul></ul><ul><ul><li>called a , </li></ul></ul><ul><li>This vector can represent any vector quantity. </li></ul>
    11. 11. Vectors <ul><li>The length of the line represents : </li></ul><ul><ul><li>the magnitude and , </li></ul></ul><ul><li>The direction of the line segment represents : </li></ul><ul><ul><li>the direction of the vector quantity. </li></ul></ul>
    12. 12. Adding Vectors <ul><li>Must be done geometrically </li></ul><ul><li>Use directed line segments </li></ul><ul><li>Add new vector to end of previous vector </li></ul><ul><li>Resultant vector is found by drawing a directed line segment from the starting point to the end of the last vector </li></ul>
    13. 13. Adding Vectors <ul><li>What does 6 + 8 equal when added by vectors? </li></ul>
    14. 14. TRY EXAMPLE 1
    15. 15. Solution <ul><li>Step 1 - Determine scale: </li></ul><ul><ul><li>1cm = 1m  </li></ul></ul><ul><li>Step 2 - Determine reference direction: </li></ul><ul><ul><li>North is directly up the page </li></ul></ul><ul><li>Step 3 - Determine a starting point: </li></ul><ul><ul><li>It appears that the person will be north of their starting point so I will start the diagram lower down the page and to the right </li></ul></ul>
    16. 16. Solution <ul><li>Step 4 - Draw the vectors given ‘top to tail’ </li></ul> 
    17. 17. Solution <ul><li>Step 5 - Draw the resultant vector: </li></ul>
    18. 18. Solution <ul><li>Step 6 - Determine the magnitude and the direction of the vector: </li></ul><ul><ul><li>This can be done either by </li></ul></ul><ul><ul><ul><li>1.  Using a ruler and a geoliner, or </li></ul></ul></ul><ul><ul><ul><li>2.  Using Pythagoras’ theorem </li></ul></ul></ul><ul><ul><ul><ul><li>R 2 = 10 2 + 15 2 </li></ul></ul></ul></ul><ul><ul><ul><ul><li>R =  10 2 + 15 2 </li></ul></ul></ul></ul><ul><ul><ul><ul><li>R =  100 + 225 </li></ul></ul></ul></ul><ul><ul><ul><ul><li>R =  325 </li></ul></ul></ul></ul><ul><ul><ul><ul><li>R = 18 m </li></ul></ul></ul></ul>
    19. 19. Solution <ul><li>To determine the direction: </li></ul><ul><li>Angle in relation to true north is: </li></ul><ul><li>360 - 56.3 = 303.7 o T </li></ul>
    20. 20. Solution <ul><li>The person has moved 18m in a direction of 304 o T from their starting position (to 3 sig. figs.) </li></ul><ul><li>Note: </li></ul><ul><ul><li>Y ou are only expected to be able to solve problems using scale diagrams, Pythagoras’ theorem and basic trigonometry </li></ul></ul>
    21. 21. Subtracting Vectors <ul><li>The vector (eg x ) that is to be subtracted, has the same magnitude but is opposite in direction (eg - x ) </li></ul>
    22. 22. TRY EXAMPLE 2 <ul><li>y - x </li></ul>
    23. 23. Solution
    24. 24. Multiplying a Vector by a Scalar <ul><li>Multiplication is simply repeated addition </li></ul><ul><li>3 x is simply x + x + x </li></ul>
    25. 25. Multiplying a Vector by a Scalar <ul><li>Also ½ x </li></ul>
    26. 26. TRY EXAMPLE 3
    27. 27. Solution <ul><li>The plane will have to fly such that the wind is one component of its velocity v w , the resultant velocity is directed N20°W v and the other component of its motion , v p , has a magnitude of </li></ul><ul><li>200 ms -1 </li></ul>
    28. 28. Solution <ul><li>Now v = v p + v w </li></ul>
    29. 29. Solution – Scale Diagram <ul><li>Draw v w as a line 2 cm long to the east. The scale is 1 cm  25 ms -1 . </li></ul><ul><li>Draw in a line 110 o anticlockwise from , v w . This represents the resultant, v . </li></ul><ul><li>Draw a line 8 cm long ( v p ) starting on the beginning of v w and intersecting v . (In practice, compasses can be used to find where the 8 cm line cuts v . </li></ul>
    30. 30. Solution – Scale Diagram <ul><li>The direction of v p can now be determined from the accurate vector diagram. Here, a ruler and protractor are required </li></ul><ul><li>Hence, v p is directed at 23 o west of north i. e . N34 o W . (Or 326 o T ) </li></ul>
    31. 31. Solution – Trigonometry Method <ul><li>From the diagram, the angle  needs to be calculated </li></ul><ul><li>  </li></ul><ul><li>  </li></ul>
    32. 32. 1-D Kinematics <ul><li>Converting to S.I. units: </li></ul><ul><li>100 km h -1 </li></ul>= 27.8 m s -1
    33. 33. 1-D Kinematics <ul><li>where v = average velocity (This is a vector quantity) </li></ul><ul><li>s = change in displacement </li></ul><ul><li> t = change in time </li></ul>
    34. 34. Kinematic Equations <ul><li>v f = v o + a t (no s ) </li></ul><ul><li>v f 2 – v o 2 = 2 as (no t ) </li></ul><ul><li>s = v o t + ½ a t 2 (no v f ) </li></ul><ul><li>s = v t t – ½ a t 2 (no v o ) </li></ul>(no a )
    35. 35. Newton <ul><li>Newton is an English Scientist who worked in this area. </li></ul><ul><li>He is the most famous scientist in the world today: </li></ul><ul><ul><li>Remember the apple falling on his head? </li></ul></ul><ul><ul><li>Besides Albert Einstein. </li></ul></ul><ul><ul><li>He lived during the 1600s </li></ul></ul>
    36. 36. Newton
    37. 37. Newton <ul><li>Science is not exact, mistakes are made. </li></ul><ul><li>Newton found mistakes in his own work after it was published. </li></ul><ul><li>He then corrected the mistakes in his personal copy. </li></ul>
    38. 38. Newton
    39. 39. Newton’s First Law <ul><li>Every body continues in its state of rest or uniform speed in a straight line unless it is compelled to change that state by a net force acting on it </li></ul><ul><li>Law of inertia </li></ul>
    40. 40. Newton’s First Law
    41. 41. Newton’s First Law
    42. 42. Newton’s First Law <ul><li>A body is resistant to change. </li></ul><ul><li>This resistance of a body to change is called inertia . </li></ul><ul><li>Does this hold for a stationary object? </li></ul><ul><li>A stationary object has zero velocity : </li></ul><ul><ul><li>and so it will remain at zero. </li></ul></ul>
    43. 43. Newton’s First Law <ul><li>Plates on a tablecloth are at rest. </li></ul><ul><li>If you pull the tablecloth quickly enough from under the plates : </li></ul><ul><ul><li>t he small and brief force of friction , </li></ul></ul><ul><ul><li>between the plates and the tablecloth is , </li></ul></ul><ul><ul><li>not significant enough to , </li></ul></ul><ul><ul><li>appreciably move the dishes. </li></ul></ul>
    44. 44. Newton’s First Law <ul><li>Does this hold for tug of war contest? </li></ul><ul><li>When the two sides are even : </li></ul><ul><ul><li>the forces are equal in magnitude and , </li></ul></ul><ul><ul><li>opposite direction. </li></ul></ul><ul><li>The resultant force will be zero. </li></ul><ul><li>This implies that the two sides will not move. </li></ul>
    45. 45. Newton’s First Law <ul><li>Does this apply to a car moving on a straight road ; </li></ul><ul><ul><li>at a constant velocity? </li></ul></ul><ul><li>The car will slow to a stop unless : </li></ul><ul><ul><li>the force applied by the engine continues. </li></ul></ul>
    46. 46. Newton’s First Law <ul><li>This appears to disobey the law : </li></ul><ul><ul><li>until friction is taken into account. </li></ul></ul><ul><li>We can ignore gravity because : </li></ul><ul><ul><li>it is acting in a vertical direction. </li></ul></ul><ul><li>Friction however is a retarding force acting : </li></ul><ul><ul><li>in the opposite direction to the motion. </li></ul></ul>
    47. 47. Newton’s First Law <ul><li>This is an example of translational equilibrium. </li></ul><ul><li>Two forces are being applied in opposite directions. </li></ul><ul><li>As the forces cancel each other out, </li></ul><ul><ul><li>no acceleration occurs. </li></ul></ul><ul><li>The object continues to move at: </li></ul><ul><ul><li>a constant velocity. </li></ul></ul>
    48. 48. Newton’s Second Law <ul><li>The acceleration of an object is directly proportional to the net force acting on it and is inversely proportional to its mass. The direction of the acceleration is in the direction of the applied net force </li></ul><ul><li>( F = m a ) </li></ul>
    49. 49. Newton’s Third Law <ul><li>Whenever one object exerts a force on a second object, the second exerts an equal and opposite on the first . </li></ul>
    50. 50. Newton’s Third Law <ul><li>Forces occur in pairs in every interaction. </li></ul><ul><li>Use your foot to push against the floor to move, </li></ul><ul><ul><li>The floor pushes back. </li></ul></ul><ul><li>You push backwards against the water in a pool, </li></ul><ul><ul><li>The water pushes back and moves you forward. </li></ul></ul>
    51. 51. Newton’s Third Law <ul><li>What happens if the force pair is not obvious? </li></ul><ul><li>E.g., a rock falling down a cliff? </li></ul><ul><ul><li>Earth pulls rock. </li></ul></ul><ul><ul><li>Rock pulls earth. </li></ul></ul>
    52. 52. Newton’s Third Law <ul><li>E.g., Car accelerating from a stop light? </li></ul><ul><ul><li>Tyre pushes on road. </li></ul></ul><ul><ul><li>Road pushes on tyre. </li></ul></ul>
    53. 53. Newton’s Third Law <ul><li>Rifle and bullet: </li></ul><ul><ul><li>Rifle pushes bullet out of barrel. </li></ul></ul><ul><ul><li>Bullet pushes rifle back into the shoulder, </li></ul></ul><ul><ul><ul><li>Kickback. </li></ul></ul></ul>
    54. 54. Newton’s Third Law <ul><li>Statue sitting on a table </li></ul>
    55. 55. Newton’s Third Law <ul><li>Be careful choosing the action/reaction pairs. </li></ul><ul><li>F N ’ is the reaction force to F N </li></ul><ul><li>F N ’ = Force exerted on the table by the statue </li></ul><ul><li>The reaction force to F g is not shown. </li></ul>
    56. 56. Newton’s Third Law <ul><li>Consider Newton’s Second Law. </li></ul><ul><li>As mass of bullet is small: </li></ul><ul><ul><li>Acceleration of bullet is large. </li></ul></ul><ul><li>As mass of rifle is relatively large: </li></ul><ul><ul><li>Acceleration of rifle is small. </li></ul></ul>
    57. 57. Newton’s Third Law <ul><li>E.g., 747. </li></ul><ul><li>747 moves by exhaust gases pushing on atmosphere? </li></ul><ul><li>What happens with a rocket in space? </li></ul><ul><ul><li>Gases push against rocket. </li></ul></ul><ul><ul><li>Rocket pushes against gas and moves forward. </li></ul></ul>
    58. 58. Newton’s Third Law <ul><li>E.g., Wagging tail of dog. </li></ul><ul><ul><li>Tail wags dog. </li></ul></ul><ul><ul><ul><li>Most noticeable when tail has a large relative mass. </li></ul></ul></ul>
    59. 59. Newton’s Third Law <ul><li>Why don’t forces cancel? </li></ul><ul><li>Kick football: </li></ul><ul><ul><li>it accelerates away from you. </li></ul></ul><ul><li>There is only one force: </li></ul><ul><ul><li>By NII, if force applied, object accelerates. </li></ul></ul><ul><li>What about equal and opposite force? </li></ul><ul><ul><li>It doesn’t act on the ball: </li></ul></ul><ul><ul><ul><li>It acts on your foot. </li></ul></ul></ul>
    60. 60. Newton’s Third Law <ul><li>Two people kicking the same soccer ball with equal but opposite force? </li></ul><ul><li>Two forces on the ball. </li></ul><ul><ul><li>Cancel each other out. </li></ul></ul>
    61. 61. Components of Projectile Motion <ul><li>The motion is in two dimensions </li></ul><ul><ul><li>H orizontal (perpendicular to the gravitational field) </li></ul></ul><ul><ul><li>V ertical </li></ul></ul><ul><ul><li>(parallel to the gravitational field) </li></ul></ul>
    62. 62. Components of Projectile Motion
    63. 63. Components of Projectile Motion <ul><li>This shows the: </li></ul><ul><ul><li>Horizontal comp is constant </li></ul></ul><ul><ul><li>Vertical component accelerates at </li></ul></ul><ul><ul><li> 9.8 m s - 2 vertically downward </li></ul></ul>
    64. 64. Components of Projectile Motion <ul><li>Motion of the two components is independent of each other. </li></ul><ul><li>Path of the projectile is parabolic. </li></ul>
    65. 65. Components of Projectile Motion <ul><li>If a monkey was in a tree and a zoo keeper wanted to fire a banana to the monkey. </li></ul><ul><li>The monkey lets go of the branch at the same time as the banana is fired. </li></ul><ul><li>Where would he aim if there was no gravity? </li></ul>
    66. 66. Components of Projectile Motion
    67. 67. Components of Projectile Motion <ul><li>Where should the zoo keeper aim if there was gravity and the monkey continues to let go of the branch? </li></ul>
    68. 68. Components of Projectile Motion
    69. 69. Components of Projectile Motion <ul><li>Both the banana and the monkey accelerate at the same rate downwards. </li></ul><ul><li>Both fall the same amount below their gravity free path. </li></ul><ul><li>Banana passes over the monkey’s head. </li></ul><ul><li>Passes over by the same amount as it was originally aimed over the monkey’s head. </li></ul>
    70. 70. Components of Projectile Motion <ul><li>What happens if you aim at the monkey? </li></ul>
    71. 71. Components of Projectile Motion <ul><li>What happens if the banana is fired slowly? </li></ul>
    72. 72. Firing Projectiles Horizontally <ul><li>What happens when you fire a projectile that has only a horizontal component? </li></ul>
    73. 73. Firing Projectiles Horizontally <ul><li>It gains a vertical component due to the force of gravity. </li></ul><ul><li>Watch the animation again and determine the value of g and the launch speed. </li></ul>
    74. 74. Firing Projectiles Horizontally
    75. 75. Dropping Vertically <ul><li>What happens if you drop a parcel from a plane? </li></ul>
    76. 76. Dropping Vertically <ul><li>Notice the horizontal components of the plane and the parcel remain equal. </li></ul><ul><li>The parcel accelerates vertically under the force of gravity. </li></ul>
    77. 77. Firing Vertically Upwards <ul><li>What happens if you fire an object vertically upwards from a moving vehicle? </li></ul>
    78. 78. Firing Projectiles Non - Horizontally <ul><li>What happens if the canon ball is fired above the horizontal? </li></ul><ul><li>Check the speeds of the components and the value of g . </li></ul>
    79. 79. Firing Projectiles Non - Horizontally
    80. 80. Firing Projectiles <ul><li>Projectiles. </li></ul><ul><li>Projectile Motion inc Calculations </li></ul>
    81. 81. Determining Characteristics of Projectiles <ul><li>How do we calculate the components speed? </li></ul><ul><li>The actual speed? </li></ul><ul><li>The range? </li></ul><ul><li>The time of flight? </li></ul>
    82. 82. Determining Characteristics of Projectiles <ul><li>Step 1 </li></ul><ul><ul><li>Determining Initial Components </li></ul></ul>
    83. 83. Determining Characteristics of Projectiles <ul><li>Horizontal Component </li></ul><ul><li> </li></ul>v i H = v i cos  Vertical Component v i V = v i sin 
    84. 84. Determining Characteristics of Projectiles <ul><li>Step 2 </li></ul><ul><ul><li>Determining Time to Maximum Height </li></ul></ul><ul><li>Note: Only vertical component affects the height.  </li></ul>
    85. 85. Determining Characteristics of Projectiles <ul><li>v fv = v ov + at </li></ul><ul><ul><li>a = -9.8ms -2 (i.e. acceleration is in the opposite direction to the motion when the projectile is still climbing)  </li></ul></ul><ul><li>0 = v o V + at </li></ul><ul><ul><li>v v = 0 (at maximum height) </li></ul></ul>
    86. 86. Determining Characteristics of Projectiles <ul><li>Step 3 </li></ul><ul><ul><li>Determining Maximum Height </li></ul></ul><ul><li>Note: Only vertical component affects the height </li></ul><ul><li>s v = v o V t + ½ at 2 </li></ul><ul><ul><li>a = -9.8ms -2 (i.e. acceleration is in the opposite direction to the motion) </li></ul></ul><ul><li>Use t from S tep 2 </li></ul>
    87. 87. Determining Characteristics of Projectiles <ul><li>Step 4 </li></ul><ul><ul><li>Determining range </li></ul></ul><ul><li>Note: Only horizontal component affects range. If the ground is flat, the time in the air = 2 x time to maximum height </li></ul><ul><li>s h = v oH t + ½ at 2 </li></ul><ul><ul><li>a = 0 in horizontal component </li></ul></ul><ul><li>s h = v oH t </li></ul><ul><ul><li>t = 2 x value of t in S tep 2 </li></ul></ul>
    88. 88. Determining Characteristics of Projectiles <ul><li>S tep 5 </li></ul><ul><li>Determining Position at Any Time </li></ul><ul><ul><li>Horizontal component: </li></ul></ul>s h = v i H t This gives distance down the range
    89. 89. Determining Characteristics of Projectiles <ul><li>Vertical component: s v = v tV t + ½ at 2 a = -9.8 ms -2 </li></ul><ul><li>This gives distance above ground. (i.e. acceleration is in the opposite direction to the motion) </li></ul>
    90. 90. Determining Characteristics of Projectiles <ul><li>You are expected to be able to determine the magnitude and direction of a velocity vector at any instant by calculating the horizontal and vertical components by using trigonometric ratios or scale a diagram </li></ul>
    91. 91. Maximum Range <ul><li>What angle gives you the maximum range? </li></ul>
    92. 92. Different Launch Height <ul><li>The final height may be different from the initial height . </li></ul><ul><li>How does this change the characteristics of flight? </li></ul><ul><ul><li>The object will still follow a parabolic path. </li></ul></ul><ul><ul><li>It will travel further. </li></ul></ul><ul><ul><li>It will drop further vertically with each unit of time than if launched at the same height. </li></ul></ul>
    93. 93. Different Launch Height <ul><li>If you are throwing a ball from shoulder height which is going to land on the ground: </li></ul><ul><ul><li>45 o is no longer the best launch angle. </li></ul></ul><ul><ul><li>A shallower angle is better. </li></ul></ul><ul><ul><li>See Parham </li></ul></ul>
    94. 94. Effect of Air Resistance <ul><li>Most projectiles do not follow a perfect parabolic path as there is another force besides gravity that acts on the projectile. </li></ul><ul><li>This force is due to the medium it travels through. In most cases, this is air. </li></ul>
    95. 95. Effect of Air Resistance <ul><li>Air is a retarding force and so resists the motion. </li></ul><ul><li>Retardation depends on the size, shape and mass of the object. </li></ul><ul><li>A large surface area will result in greater air resistance effects. </li></ul><ul><li>A streamlined ‘bullet’ shape will minimise the effect of air resistance. </li></ul>
    96. 96. Effect of Air Resistance <ul><li>Air is a retarding force and so resists the motion. </li></ul><ul><li>Retardation depends on the speed, surface texture of the object. </li></ul><ul><li>Retardation depends on air density. </li></ul><ul><li>A streamlined ‘bullet’ shape will minimise the effect of air resistance. </li></ul>
    97. 97. Effect of Air Resistance <ul><li>The overall effects cause projectiles to travel slower, to take longer in flight and not to travel as high or as far. </li></ul>
    98. 98. TRY EXAMPLE 4
    99. 99. Solution
    100. 100. Solution – Part (a) <ul><li>s = v o t + ½ a t 2 </li></ul><ul><li>Using Pythagoras’ Theorem: </li></ul><ul><li>s 2 = ( v o t ) 2 + ( ½ a t 2 ) 2 </li></ul><ul><li>s 2 = (10 x 3) 2 +(½ x 9.8 x 3 2 ) 2 </li></ul><ul><li>s =  900 + 1945 </li></ul><ul><li>s =  2845 </li></ul><ul><li>s = 53.3 m </li></ul>1/2
    101. 101. Solution – Part (a) <ul><li>  </li></ul><ul><li>tan  = 44.1/30 </li></ul><ul><li> = 55.8 o </li></ul><ul><li>s = 55.3 m at 55.8 o down from horizontal </li></ul>1/2
    102. 102. Solution – Part (a) <ul><li>v f = v o + a t </li></ul><ul><li>Using Pythagoras’ Theorem: </li></ul><ul><li>( v f ) 2 = ( v o ) 2 + ( a t ) 2 </li></ul><ul><li>( v f ) 2 = 10 2 + (9.8 x 3) 2 </li></ul><ul><li>( v f ) 2 = 100 + 864 </li></ul><ul><li>( v f ) 2 = 964 </li></ul><ul><li>v f = 31.0 ms -1 </li></ul>
    103. 103. Solution – Part (a) <ul><li>  </li></ul><ul><li>tan  = 29.4/10 </li></ul><ul><li> = 71.2 o </li></ul><ul><li>v t = 31.0 ms -1 at 71.2 o down from horizontal </li></ul>
    104. 104. Solution – Part (b) <ul><li>O nly the vertical component of the motion need be considered </li></ul><ul><li>Here, v oV = 0 and a V = 9.8 ms -2 down </li></ul><ul><li>Hence, s V = 122.5 m down </li></ul><ul><li>s V = v o V t + ½ a V t 2 </li></ul><ul><li>122.5 =0 + ½ x 9.8 x t 2 </li></ul><ul><li>t 2 = 122.5/4.9 </li></ul><ul><li>t =  25 </li></ul><ul><li>t = 5 s </li></ul>
    105. 105. TRY EXAMPLE 5
    106. 106. Solution – Part (a)
    107. 107. Solution – Part (a) <ul><li>v oH = v o cos    </li></ul><ul><li>v oH = 98 x cos 30 o </li></ul><ul><li>v oH = 84.9 ms -1 horiz </li></ul><ul><li>v o V = v o sin  </li></ul><ul><li>v o V = 98 x sin 30 o </li></ul><ul><li>v o V = 49 ms -1 vert </li></ul>
    108. 108. Solution – Part (b) <ul><li>At the top of the motion, v v = 0   </li></ul><ul><li>v fV 2 – v oV 2 = 2 as V   </li></ul><ul><li>0 - 49 2 = 2 x -9.8 x s V </li></ul><ul><li>s V = -2401/-19.6 </li></ul><ul><li>s V = 122.5 m </li></ul>
    109. 109. Solution – Part (c) <ul><li>v o V = - v V . </li></ul>
    110. 110. Solution – Part (d) <ul><li>The acceleration at the top of the flight is 9.8 ms -2 downward due to gravity. </li></ul>
    111. 111. Solution – Part (e) <ul><li>Consider the horizontal component of the motion. </li></ul><ul><li>s H = v o H t + a H t 2 </li></ul><ul><li>s H = 84.9 x 10 + 0 </li></ul><ul><li>s H = Range = 849 m </li></ul>
    112. 112. Solution – Part (f) <ul><li>v t = v o + a t </li></ul><ul><li>Triangle is equilateral </li></ul><ul><li>The third side </li></ul><ul><li>= 98 ms -1 </li></ul><ul><li>A ll angles are 60 o </li></ul><ul><li>v = 98 ms -1 30 o down from horizontal </li></ul>