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PROJECTILES Motion in Two Dimensions  Copyright Sautter 2011
Please let me promote some of my  other work on the next slide and  then the selected presentation  will start. Thank you. Walt S.
I I have written six books: "Sticks - A Golfer's Tale" - A hacker's dream comes true. "Fish Farm" - Revenge of the old people. "Coach" - A mystery set a rural, bigoted, nineteen fifties football town.  "The Three Dollar Phoenix" - A mystery set in Newark, New Jersey in the 1970s. "The Divine Comedy MMIX" - A humorous play about Jesus returning. "The Blood of Judas" - A horror story of revenge set in Nazi Germany. All are available at :  smashwords .com or at :  wsautter .com I have video trailers for "Coach", "Fish Farm" and "The Blood of Judas" at:  youtube .com Please take a look. Thanks. Walt Sautter -  wsautter @ optonline .net PS - Lots more stuff besides books at:  wsautter .com
Motion in Two Dimension - Projectiles ,[object Object],[object Object],[object Object],[object Object]
Projectile Motion JUST AFTER FIRING S y  =0, S x  = 0 V y  = +max V x  = constant A y  = g, A x  =0 AT HIGH POINT S y  = max, S x  = ½ max V y  = 0, V x  = constant A y  = g, A x  =0 JUST BEFORE  LANDING S y  = 0, S x  = max V y  = -max V x  = constant A y  = g, A x  =0 CLICK HERE
A X COMPONENT Y COMPONENT X COMPONENT Y COMPONENT B Y COMPONENT X COMPONENT C Vector Components
S y V y inst  = slope of tangents V y Time V y  = 0 A y inst  = slope of tangents A y Time A y inst  = - 9.8 meters / sec 2 Projectile Motion (vertical component) Time
V o S x Path of object Without gravity S y  = actual height  of object Distance fallen due to gravity (1/2 g t 2 ) Vertical height if gravity did not  act on the object (V o  sin    t)  S y  = height without effects of gravity – distance fallen due to gravity S y  = V o  sin     t  +  ½ g t 2   (the value of gravity is negative) Horizontal distance traveled by projectile  S x  = V o  cos    t Vertical & Horizontal Displacements of a Projectile Actual projectile path Object Projected at Angle  
Water spraying from a hose is a common example of projectile motion The  vertical motion  of the water is  accelerated by gravity. The  horizontal motion  is  constant   velocity  and is unaffected by gravity! Velocity Components of Projectile Motion V y  = V o  sin  + gt V x  = V o  cos  
Vertical displacements are equal for projectiles and dropped objects  however horizontal displacements are greater for projectiles.  Vertical acceleration for both dropped objects and projectiles is that  of gravity (-32 ft/s 2 , -9.8 m/s 2 ). They both hit the ground at the exact same time but, of course, the projectile is further away ! Accelerated by gravity Projectile V x  > 0 Constant velocity Vertical & Horizontal Motion  of a Projectile vs. a Dropped Object Dropped Object V x  = 0
Vertical & Horizontal Components of Motion during Projectile Flight V y  = + max V y  = - max V y  = 0 Horizontal component of projectile motion (Constant velocity – no acceleration) Vertical component of projectile motion (Accelerated by gravity)
Projectile Motion Equations Vertical Displacement Horizontal Displacement Vertical Velocity Horizontal Velocity Vertical Acceleration Horizontal Acceleration a  =  g y a  =  o x y o S  =  V  sin  t  +  1/2  g  t 2  V  =  V  cos  x o  V  =  V  sin  +  g  t y o  S  =  V  cos  t x o 
Special Situation Projectile Equations Maximum Height Maximum Horizontal Distance (Range) h  =  -  V  sin max o 2 2 2 g -----------------  R  =  -  V  sin  2  2 ----------------- o g h max range
When an object is dropped  at the exact same time a  projectile is fired at the  falling object, aiming directly at the object always insures a direct hit. Why ? Because the object and the projectile once fired, are in both in free fall ! Projectiles in Free Fall
Solving Projectile Problems A ball is thrown horizontally at 12 m/s from a building 30 meters high. (a) How long will is be in the air? (b) How far from the base of the building will it land? ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Equations to be Used V o  = 12 m/s 30 m y o S  =  V  sin  t  +  1/2  g  t 2  S  =  V  cos  t x o 
[object Object],[object Object],Solving Projectile Problems A ball is thrown upward at a 30 0  angle, at 12 ft/s from a building 300 feet high. (a)How long will is be in the air? (b) How far from the base of the building will it land? (a) using the equation shown we insert –300 for S y , 12 for V o  30 0  for the angle and – 32 ft/s 2  for gravity (ENG) and get –300 = 12(sin 30 0  ) t + ½ (-32) t 2  which gives  –300 = 6 t – 16 t 2  or 16 t 2  –6 t  - 300 = 0 (a quadratic). Using a = +16, b = - 6 and c = - 300 we insert these values in the quadratic equation find t =  -3.21 or + 5.46 The negative t means the ball hits the ground before it is thrown in is therefore obviously  wrong. t = +5.46 sec  (b) Using the equation shown and the time value from part (a) we get V x  = 12 (cos 30 0 ) 5.46 = 56.7 feet  Equations to be Used V o  = 12 m/s 300 m 30 0 y o S  =  V  sin  t  +  1/2  g  t 2  X  +  X  +  =  y 2 a b c x  = - b  b  -  2 a + - 2 ------------------------- / ----------------------------- 4  a c S  =  V  cos  t x o 
Solving Projectile Problems A projectile is shot on level ground at a 45 degree angle with a velocity of 20 ft/s. (a) How high will it travel? (b) How far will it go? ,[object Object],[object Object],Equations to be Used (a) using the equation shown we insert 20 for V o  45 0  for the angle and – 32 ft/s 2  for gravity (ENG) and get h max  = - ((20) 2  (sin 45 0 ) 2  )/ (2 (-32))   which gives (400 x (0.707 )2 ) / -64  .  Maximum height = 3.13 feet. (b) Using the equation shown we get R = - ((20) 2  x (sin 2 x 45 0 )) / - 32 = 12.5 feet  h max R h  = -  V  sin max o 2 2 2 g -----------------  R  = -  V  sin  2  2 ----------------- o g
Solving Projectile Problems A ball is throw downward at an angle of 30 0  with a velocity of 10 m/s from a building 40 meters high. What is the velocity of the ball when it hits the ground? ,[object Object],[object Object],[object Object],Using the given equations, V o  = 10 m/s,  g = - -9.8 m/s 2  (MKS) and S y  = - 40 m we must first find time (t).  -40 = 10 sin (-30 0 ) t + (-9.8) t 2   , 9.8 t 2  + 5 t – 40 = 0. Solving the quadratic formula with a = 9.8,  b = 5 and c = -40 we get t = - 2.29 or + 1.78 seconds. Of course, the + 1.78  is the correct answer. To find the vertical velocity, V y  = 10 (sin – 30 0 ) + (-9.8) 1.78, V y  = -22.5 m/s  To find the horizontal velocity, V x  = 10 (cos –30 0 ) 1.78 = +15.4 m/s V o  = -10 m/s 40 m -30 0 V  =  V  sin  +  g  t y o  V  =  V  cos  x o  y o S  =  V  sin  t  +  1/2  g  t 2 
Previous Problem (continued) ,[object Object],[object Object],[object Object],[object Object],V y  = -22.5 m/s   V x  = +15.4 m/s V resulting
[object Object],Solving Projectile Problems ,[object Object],[object Object],Check - At the highest point, V y  =0 and 0 = 20 (sin 45 0 ) + (-32) t = 0.442 seconds at the highest point and when descending t must be larger than 0.442 and smaller than 0.884 (2 x 0.442), the  time when the object lands. The answer is between these two values! V o  = 20 45 0 V = - 5, t = ? V  =  V  sin  +  g  t y o 
Now it's time for you to try  some problems on your own ! The problems are similar to the ones which have been  solved so look back and review the appropriate problem if you get stuck !
[object Object],[object Object],[object Object],Click Here For answers A ball is thrown at 40 0  above the horizontal at 4.0 m/s. What is its horizontal velocity after 0.50 seconds? (A) 2.6 m/s  (B) 3.1 m/s  (C) 3.4 m/s (D) 5.5 m/s What is the vertical velocity in the previous problem After 0.50 seconds ? (A) 2.6 m/s  (B) 3.1 m/s  (C) 4.9 m/s (D) 7.5 m/s What must be the original velocity of a projectile in order to reach a target 90 km away ? (A) 320 m/s  (B) 939 m/s  (C) 882 m/s (D) none Gravity  is its  acceleration It is -9.8  m /s 2 (MKS units) A ball is thrown at a 30 0  angle at 10 m/s. How far on the horizontal will it land ? (A) 9.8 m  (B) 0.88 ft  (C) 8.8 m (D) 10.2 m  Hint: use The vector Sum of V x  + V y Hint: maximum horizontal  distance occurs at an angle of 45 0
Real Projectile Motion ,[object Object],[object Object],[object Object],[object Object],[object Object]
Ideal vs. Real Projectile Paths Projectile Path without Air Resistance Projectile Path with Air Resistance
The End

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Projectiles

  • 1. PROJECTILES Motion in Two Dimensions Copyright Sautter 2011
  • 2. Please let me promote some of my other work on the next slide and then the selected presentation will start. Thank you. Walt S.
  • 3. I I have written six books: "Sticks - A Golfer's Tale" - A hacker's dream comes true. "Fish Farm" - Revenge of the old people. "Coach" - A mystery set a rural, bigoted, nineteen fifties football town. "The Three Dollar Phoenix" - A mystery set in Newark, New Jersey in the 1970s. "The Divine Comedy MMIX" - A humorous play about Jesus returning. "The Blood of Judas" - A horror story of revenge set in Nazi Germany. All are available at : smashwords .com or at : wsautter .com I have video trailers for "Coach", "Fish Farm" and "The Blood of Judas" at: youtube .com Please take a look. Thanks. Walt Sautter - wsautter @ optonline .net PS - Lots more stuff besides books at: wsautter .com
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  • 5. Projectile Motion JUST AFTER FIRING S y =0, S x = 0 V y = +max V x = constant A y = g, A x =0 AT HIGH POINT S y = max, S x = ½ max V y = 0, V x = constant A y = g, A x =0 JUST BEFORE LANDING S y = 0, S x = max V y = -max V x = constant A y = g, A x =0 CLICK HERE
  • 6. A X COMPONENT Y COMPONENT X COMPONENT Y COMPONENT B Y COMPONENT X COMPONENT C Vector Components
  • 7. S y V y inst = slope of tangents V y Time V y = 0 A y inst = slope of tangents A y Time A y inst = - 9.8 meters / sec 2 Projectile Motion (vertical component) Time
  • 8. V o S x Path of object Without gravity S y = actual height of object Distance fallen due to gravity (1/2 g t 2 ) Vertical height if gravity did not act on the object (V o sin  t)  S y = height without effects of gravity – distance fallen due to gravity S y = V o sin  t + ½ g t 2 (the value of gravity is negative) Horizontal distance traveled by projectile S x = V o cos  t Vertical & Horizontal Displacements of a Projectile Actual projectile path Object Projected at Angle 
  • 9. Water spraying from a hose is a common example of projectile motion The vertical motion of the water is accelerated by gravity. The horizontal motion is constant velocity and is unaffected by gravity! Velocity Components of Projectile Motion V y = V o sin + gt V x = V o cos  
  • 10. Vertical displacements are equal for projectiles and dropped objects however horizontal displacements are greater for projectiles. Vertical acceleration for both dropped objects and projectiles is that of gravity (-32 ft/s 2 , -9.8 m/s 2 ). They both hit the ground at the exact same time but, of course, the projectile is further away ! Accelerated by gravity Projectile V x > 0 Constant velocity Vertical & Horizontal Motion of a Projectile vs. a Dropped Object Dropped Object V x = 0
  • 11. Vertical & Horizontal Components of Motion during Projectile Flight V y = + max V y = - max V y = 0 Horizontal component of projectile motion (Constant velocity – no acceleration) Vertical component of projectile motion (Accelerated by gravity)
  • 12. Projectile Motion Equations Vertical Displacement Horizontal Displacement Vertical Velocity Horizontal Velocity Vertical Acceleration Horizontal Acceleration a = g y a = o x y o S = V sin t + 1/2 g t 2  V = V cos x o  V = V sin + g t y o  S = V cos t x o 
  • 13. Special Situation Projectile Equations Maximum Height Maximum Horizontal Distance (Range) h = - V sin max o 2 2 2 g -----------------  R = - V sin 2  2 ----------------- o g h max range
  • 14. When an object is dropped at the exact same time a projectile is fired at the falling object, aiming directly at the object always insures a direct hit. Why ? Because the object and the projectile once fired, are in both in free fall ! Projectiles in Free Fall
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  • 21. Now it's time for you to try some problems on your own ! The problems are similar to the ones which have been solved so look back and review the appropriate problem if you get stuck !
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  • 24. Ideal vs. Real Projectile Paths Projectile Path without Air Resistance Projectile Path with Air Resistance