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Objects Thrown Vertically Upward

Objects Thrown Vertically Upward. Objects Thrown Vertically Upwards. Upward motion and downward motion are perfectly symmetrical A motion path upward, including speed at every point along that path, is an exact copy of the motion path downward

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Objects Thrown Vertically Upward

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  1. Objects Thrown Vertically Upward

  2. Objects Thrown Vertically Upwards • Upward motion and downward motion are perfectly symmetrical • A motion path upward, including speed at every point along that path, is an exact copy of the motion path downward • If a ball is projected upward with a velocity of +30 m/s, it will have a downward velocity of-30 m/s when it returns to the same height

  3. Objects Thrown Vertically Upwards • Objects slow down as they rise upwards • Start off with some initial positive velocity • Acceleration is g (-10 m/s/s) • At the peak, velocity is 0 m/s • Time Up = Time Down • Hang Time = Total Time in the Air

  4. Objects Thrown Vertically Upwards

  5. Position vs. Time Graph • Curved line; indicates accelerated motion • Slope of position-time graph is velocity

  6. Position vs. Time Graph Time Up = 4 s Slope is zero; stops moving Slopes are - and getting more -; speeding up in - direction Slopes are + and getting less +; slowing down in + direction Hang Time = 8 s Time Down = 4 s

  7. Velocity vs. Time Graph • Straight, diagonal line; shows acceleration • Slope of velocity-time graph is acceleration • Constant, negative slope • Indicates a constant, negative acceleration

  8. Velocity vs. Time Graph Moving Upward & Slowing Down Initial Velocity = +40 m/s Moving Downward & Speeding Up Final Velocity= -40 m/s Stops Moving

  9. Example #1 Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height.

  10. Example #1 Givens: vi = 26.2 m/s vf = 0 m/s a = -10 m/s2 Find: d = ?? Use: vf2 = vi2+ 2 a d (0 m/s)2 = (26.2 m/s)2 + 2 (-10 m/s2) d d = 34.3 m

  11. Example #2 A tennis ball is thrown straight up with an initial velocity of +22.5 m/s. It is caught at the same distance above the ground from which it was thrown. (a) How high does the ball rise? (b) How long does the ball remain in the air?

  12. Example #2 (a) Givens: vi = +22.5 m/s a = -10 m/s2 vf = vtop = 0 m/s Find: d = ?? • (b) Givens: • vi = +22.5 m/s • vf = -22.5 m/s • a = -10 m/s2 • d = 0 m • Find: t = ??

  13. Example #2 (a) Use: vf2= vi2+ 2 a d (0 m/s)2 = (22.5 m/s)2 + 2 (-10 m/s2) d 0 m2/s2 = 506 m2/s2 + (-20 m/s2) d (20 m/s2) d = 506 m2/s2 d = 25.3 m (b) Use: vf= vi+ a t -22.5 m/s= 22.5 m/s + (-10 m/s2) t -45.0 m/s = (-10 m/s) t t = 4.50 s

  14. Example #3 A ball is thrown vertically upward from a height of 0.500 m with an initial velocity of 25.0 m/s. (a) What is the maximum height that the ball will reach? (b) How long will it take for the ball to reach its maximum height?

  15. Example #3 Givens: vi = +25.0 m/s a = -10 m/s2 vf = vtop = 0 m/s Find: (a) d = ?? (b) t = ??

  16. Example #3 (a) Use: vf2= vi2+ 2 a d (0 m/s)2 = (25.0 m/s)2 + 2 (-10 m/s2) d d = 31.3 m Remember: the ball started 0.500 m off the ground, so… d = 31.3 m + 0.500 m d = 31.8 m (b) Use: vf= vi+ a t 0 m/s= 25.0 m/s + (-10 m/s2) t t = 2.50 s

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