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Physics 2053C – Fall 2001

Physics 2053C – Fall 2001. Chapter 3 Motion in Two Dimensions Kinematics. Horizontal Motion. Vertical Motion. X = X 0 + V 0x t V x = V 0x. Y = Y 0 + V 0y t – ½ g t 2 V y = V 0y - g t. Motion in Two Dimensions. Motion in the horizontal and vertical directions are independent.

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Physics 2053C – Fall 2001

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  1. Physics 2053C – Fall 2001 Chapter 3 Motion in Two Dimensions Kinematics Dr. Larry Dennis, FSU Department of Physics

  2. Horizontal Motion Vertical Motion X = X0 + V0x t Vx = V0x Y = Y0 + V0y t – ½ g t2 Vy = V0y - g t Motion in Two Dimensions • Motion in the horizontal and vertical directions are independent. • X and Y are independent (linked only by time). • There is no acceleration in the X direction. • There is acceleration in the Y direction.

  3. Example: Range • An object is thrown with velocity of 25 m/s. What is the farthest distance it can travel horizontally before hitting the ground, assuming no air resistance and level ground? (Xo = 0, Yo = 0, ay = -g and ax = 0 ) Y = Voy t – ½ g t2 used to determine the values of twhere Y = 0 and; X = Vox t can be used with the value of t to determine how far the object goes. Voy=Vosin Vox=Vocos

  4. Example: Finding Landing Time • An object is thrown with velocity of 25 m/s. What is the farthest distance it can travel horizontally before hitting the ground, assuming no air resistance and level ground? Solve for t where Y = 0: Y = 0 = Voy t – ½ g t2 0 = Voy t – ½ g t2 0 = t * ( Voy– ½ g t)  t = 0 or 0 = ( Voy– ½ g t) 0 = ( Voy– ½ g t)  t = 2Voy/g Voy=Vosin Vox=Vocos

  5. Example: Finding Horizontal Distance • An object is thrown with velocity of 25 m/s. What is the farthest distance it can travel horizontally before hitting the ground, assuming no air resistance and level ground? Starting with the time when Y = 0: t = 2Voy/g Solve for the horizontal distance: X = Vox t= Vox * (2Voy)/g = ( Vocos 2Vosin ) / g X = ( 2Vo2cos sin ) / g Voy=Vosin Vox=Vocos

  6. Finding Maximum Horizontal Distance • An object is thrown with velocity of 25 m/s. What is the farthest distance it can travel horizontally before hitting the ground, assuming no air resistance and level ground? X = ( 2Vo2cos sin )/g

  7. To find the landing time, solve for Y = 0. 0 = 29 + 20*sin(60)*t – 5 t2 = 29 + 17.3 t – 5 t2 t = ( -17.3 ± √(17.3)2 – 4 * (-5) * 29 ) / ( 2 * (-5) ) t = 4.7 s or –1.24 s Launching and Landing at Different Heights  = 60o Y = Yo + Voy t – ½ g t2 X= Vox t Vo = 20 m/s Yo = 29.0 m

  8. Launching and Landing at Different Heights  = 60o Y = Yo + Voy t – ½ g t2 X= Vox t Vo = 20 m/s Yo = 29.0 m Where does the projectile land? t = 4.7 s is the time we are interested in. Xland = Vox t = 20 * cos (60) * 4.7 = 47 m

  9. Launching and Landing at Different Heights  = 60o Y = Yo + Voy t – ½ g t2 X= Vox t Vo = 20 m/s Yo = 29.0 m How high does the projectile go?  Maximum height occurs when Vy = 0. Vy = Voy – g t  0 = Voy – g t  tmax= Voy/g = 17.3/10 = 1.73 s

  10. Launching and Landing at Different Heights  = 60o Y = Yo + Voy t – ½ g t2 X= Vox t Vo = 20 m/s Yo = 29.0 m How high does the projectile go?  Maximum height occurs when Vy = 0 and t = 1.73 s. Y = Yo + Voy t – ½ g t2 = 29.0 + 17.3 *(1.73) – 5 (1.73)2 = 44.0 m

  11. Launching and Landing at Different Heights  = 60o Vo = 20 m/s Yo = 29.0 m

  12. Review of Motion in One-dimension • Displacement • Average Velocity, Velocity, Speed • Average Acceleration • Motion with Constant Velocity • Motion with Constant Acceleration

  13. X Vave = = t Change in position (Xfinal - Xinitial ) Change in time Vave = (tfinal – tinitial ) Review of Motion in One-dimension • Average Velocity (or Speed) Note: speed = magnitude of the velocity

  14. V aave = = t Change in velocity (Vfinal - Vinitial ) Change in time aave = (tfinal – tinitial ) Review of Motion in One-dimension • Average Acceleration Note: Magnitudes are either positive or zero, but never negative.

  15. Review of Motion in One-dimension • Motion with constant velocity The acceleration is zero. X = Xo + Vo t Note: When the velocity is constant, the average speed and instantaneous speed are the same.

  16. Review of Motion in One-dimension • Motion with constant acceleration. X = Xo + Vo t + ½ a t2 V = Vo + a t V2 = Vo2 + 2 a (X – Xo) Note: When the accleration is constant, the average acceleration and instantaneous acceleration are the same.

  17. Review of Motion in One-dimension Sample Questions: Page 41 of Giancoli Questions 9, 10, 11, 12, 14 Sample Problems: Page 42 & 43 of Giancoli # 5, 19, 23, and 35 (see back of book for answers)

  18. Next Time • Quiz on Motion in one and two dimensions (primarily one dimension). • At this point you should be able to do all of the CAPA problems. • Begin Chapter 4. • Please see me with any questions or comments. See you Wednesday.

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