Lesson 6 - Freefall - general Physics - SHS

freefall discussion

1. F r e e f a l l

2. Free fall idealized motion of a falling object, which is acted upon only by the force of gravity acceleration of a free-falling body is called acceleration due to gravity denoted by "g" “g” is equal to 9.8 m/s2 at Earth's surface directed toward Earth's center.

3. Freefall In 1971 David Scott proved freefall when he dropped a feather and a hammer from the same height on the surface of the moon, where air resistance is almost absent.

4. The motion of an object that is thrown upward and which eventually returns to its starting point exhibits two symmetries

5. Time symmetry the time required for the object to reach its maximum height equals the time for it to return to its starting point.

6. Speed symmetry shows that at any displacement above the point of release, the speed of the body during the upward trip equals the speed during the downward trip . In addition, the velocity at the maximum height is equal to zero, but the acceleration is still equal to g.

7. Air resistance describes the forces that are in opposition to the relative motion of an object as it passes through the air.

8. Formula d = V i t + ½ at2 d = ½ (V i + V f ) t V f 2 = V i 2 + 2ad V f = V i + at FREE FALL PROBLEMS

9. Freefall In Free fall, speed and distance is always positive while velocity and displacement can be positive or negative. If the object goes up the velocity is positive and if the object goes down the velocity is negative.

10. Sample problem: 1. A ball dropped from a cliff. what is the speed in 7 seconds later? what is the velocity of the ball this time? how far does this travel during this time? what is the displacement of the ball?

11. 1. A ball dropped from a cliff. what is the speed in 7 seconds later? what is the velocity of the ball this time? how far does this travel during this time? what is the displacement of the ball? Given: t = 7 secs v i = 0 m/s g = 9.8 m/s 2

12. 1. A ball dropped from a cliff. what is the speed in 7 seconds later? what is the velocity of the ball this time? Given: t = 7 secs v i = 0 m/s g = 9.8 m/s 2 Formula: V f = V i + at Solution: V f = V i + at = 0 m/s + (9.8) (7) Speed is 68.6 m/s, velocity is (- 68.6 m/s)

13. A ball dropped from a cliff. (c.) how far does this travel during this time? (d) what is the displacement of the ball? Given: t = 7 secs v i = 0 m/s v f = 68.6 m/s g = 9.8 m/s 2 Formula: d = ½ (V i + V f ) t Solution: d = ½ (V i + V f ) t d = ½ ( 0 m/s + 68.6 m/s)(7) The distance travelled is 240.1 m

14. 1. A ball dropped from a cliff. how far does this travel during this time? what is the displacement of the ball? Given: t = 7 secs v i = 0 m/s g = 9.8 m/s 2 Formula: V f = V i + at Solution: V f = 0 m/s + (9.8m/s 2 )(7 s) V f = 68.6 m/s Speed is 68.6 m/s, velocity is (- 68.6 m/s)

15. FREE FALL problems

16. Formula d = V i t + ½ at 2 d = ½ (V i + V f ) t V f 2 = V i 2 + 2ad V f = V i + at Back

17. Sample problem: 1. From what height must water fall from a dam to sink the blade of the turbine with a velocity of 35 m/s? Formula

18. 1. From what height must water fall from a dam to sink the blade of the turbine with a velocity of 35 m/s? Given: v i = 0 m/s v f = 35 m/s g = - 9.8 m/s 2 Formula: V f 2 = V i 2 + 2ad Solution: d = (35 m/s) 2 – (0m/s) 2 2 (- 9.8 m/s 2 ) d = 1225 m 2 /s 2 - 19.6 m/s 2 d = - 62.5 m V f 2 - V i 2 = 2ad V f 2 - V i 2 = d 2g Formula

19. Sample problem: 2. Victor threw a ball upwards with an initial velocity of 2 m/s. He was able to catch it before it reached the ground What was the velocity after a second? What was the displacement in the 1 st second? How long did it take the ball to reach its maximum height?

20. 2. Victor threw a ball upwards with an initial velocity of 2 m/s. He was able to catch it before it reached the ground a. What was the velocity after a second? Given: t = 1 sec v i = 2 m/s g = 9.8 m/s 2 Formula: V f = V i + at Solution: V f = V i + at = 2 m/s - (9.8m/s 2 ) (1s) V f = - 7.8 m/s Formula

21. 2. Victor threw a ball upwards with an initial velocity of 2 m/s. He was able to catch it before it reached the ground b. What was the displacement in the 1st second? Given: t = 1 sec v i = 2 m/s g = - 9.8 m/s 2 Formula: d = v i t + ½ gt 2 Solution: d = (2 m/s)(1s) + ½ (-9.8 m/s) (1s) 2 = 2 m – 4.9 m d = - 2.9 m Formula

22. 2. Victor threw a ball upwards with an initial velocity of 2 m/s. He was able to catch it before it reached the ground b. What was the displacement in the 1st second? Given: t = 1 sec v i = 2 m/s v f = - 7.8m/s Formula: d = ½ (V i + V f ) t Solution: d = ½ (2 m/s -7.8 m/s) 1 sec = ½ (5.8 m/s) d = - 2.9 m Formula

23. 2. Victor threw a ball upwards with an initial velocity of 2 m/s. He was able to catch it before it reached the ground c. How long did it take the ball to reach its maximum height? Given: v i = 2 m/s v f = 0 m/s g = - 9.8 m/s Formula: v f = v i + gt t = V f - V i g Solution: t = (0) – (2m/s) (-9.8 m/s 2 ) = (-2 m/s) (-9.8 m/s 2 ) t = 0.204 s Formula

24. Sample problem: 3. A ball rolls off the building that is 56 m high. Calculate the time that it takes for the ball to hit the ground.