1. FREEFALL GROUPWORK Stay at your assigned computer Talk only to the people at your computer

2. An object is tossed 24.0 m/s upward. How high does it go?

3. Two objects start at the same place and travel 300 m. Both start from rest at the same time and reach 300m at the same time. One object travels at constant speed, and the other accelerates. If the first object has an acceleration of 4 m/s 2, what must the constant speed of the other object be?

4. What was the initial velocity of a ball thrown from a 80.0 m tall cliff, if it strikes the base of the cliff with a velocity of 62.0 m/s?

5. THINKER QUESTION: A brick is dropped from rest from the top of a 40 m tall building. On the way down it is observed passing a 1.2 m tall window in.10 s. What is the height of the bottom of the window? ? 40 m

6. Now you will check and correct your answers Raise your hand & show me your papers. I will tell you the password to check your answers

7. DISTANCE ( m ) TIME (s) 35 30 25 20 15 10 5 0 -5 -10 -15 -20 -25 -30 -35 -40 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 A.What time interval (if any) has negative velocities in positive positions? ANSWER : 32-40 s

8. An object is tossed 24.0 m/s upward. How high does it go? v i = 24 So right here I set my frame of reference…UP is positive a = -9.8 + Pos direction - Getting slower v f = 0 Goes UP until it stops! d = ? v f 2 = v i 2 + 2ad 0 = 24 2 + 2(-9.8)d 0 = 576 - 19.6d 19.6d = 576 d = 29.4 m

9. Two objects start at the same place and travel 300 m. Both start from rest at the same time and reach 300 m at the same time. One object travels at constant speed, and the other accelerates. If the first object has an acceleration of 4 m/s 2, what must the constant speed of the other object be? Constant Speed Object dtdt v = Accelerating Object v i = 0 d = 300 a = 4 t = ? d = v i t + ½ at 2 300 = ½ (4)t 2 t = 12.2 s 300 12.2 v = v = 24.6 m/s

10. What was the initial velocity of a ball thrown from a 80.0 m tall cliff, if it strikes the base of the cliff with a velocity of 62.0 m/s? vi vi = ? d = 80 so down is POS a = 9.8 vf vf = 62 v f 2 = v i 2 + 2ad 62 2 = v i 2 + 2(9.8)(80) v i = +- 47.7 m/s

11. THINKER QUESTION: A brick is dropped from rest from the top of a 40 m tall building. On the way down it is observed passing a 1.2 m tall window in.10 s. What is the height of the bottom of the window? ? 40 m Passing window: d = 1.2 a = 9.8 t =.10 v i = ? d = v i t + ½ at 2 1.2 = v i (.10) + ½ (9.8)(.10) 2 v i = 11.5 m/s From top of building: d = ? a = 9.8 v f = 11.5 v i = 0 v f 2 = v i 2 + 2ad 11.5 2 = 2(9.8)d d = 6.76 m 32.0m