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 m/s. Determine the height to which the vase will rise above its initial height.

10. Example #1 Givens: Find: d = ?? vi = 26.2 m/s vf = 0 m/s a = -10 m/s2
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 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: (b) Givens: vi = +22.5 m/s vi = +22.5 m/s
a = -10 m/s2
vf = vtop = 0 m/s
Find: d = ??
(b) Givens:
vi = m/s
vf = 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 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. Remember: the ball started 0.500 m off the ground, so…
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 m off the ground, so…
d = 31.3 m 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