Freefall

1. Freefall

2. Beginning of Period Problem: An automobile slows from 15.0 m/s to a complete stop in a distance of 15.0 m. Assuming the same braking power and conditions, how much distance is needed to stop the same car from 30.0 m/s? Solution 1: Solve for the acceleration: Given and Unknown information: vo = 15.0 m/s Dx = 15.0 m v = 0 m/s a = ?

3. Next solve for the Dx for the new speed with the same equation.

4. Solution 2: Look at the equation used, and note that there is a proportionality between the initial velocity and the distance. Substitute in the final velocity of zero right away as well. 0 Rearrange terms: Finally

5. Note: The stopping distance is proportional to the square of the initial speed. The speed for the second case is double the first case. If the initial velocity doubles, the stopping distance is 4 times as large. So, 4 times 15.0 m is 60.0 m.

6. New Topic: Freefall. The equations derived thus far are for uniformly accelerated motion. In other words, the _____________ remains ____________ for the entire motion of the object. acceleration constant This is a special case, and few real objects move with constant acceleration. There is one example of objects moving with a constant acceleration: Those objects moving under the influence of _________ near the ___________ of ____________ . gravity surface Earth Near the surface of Earth, the acceleration due to gravity has a measured value of __________ or __________ , pointing downward. 9.80 m/s2 32.2 ft/s2 Since the acceleration is constant, it will be given its own variable: g magnitude This variable will represent only the _____________ of the acceleration, not the _____________ ! direction

7. To show direction, we need a coordinate system: the vertical y-axis. negative This end will be the __________ y-axis. Anything pointing ____________ will be considered _________ . upward negative positive This other end will be the __________ y-axis. Anything pointing ____________ will be considered _________ . positive downward downward Since gravity pulls ___________ , the acceleration due to gravity also points ___________ . Thus the acceleration will be written as: a = g downward ‘g’ is the magnitude of the acceleration, and the (+) sign is the direction of the acceleration.

8. Now let us consider the other variables used. For the coordinate line, up is positive. The zero location can be defined anyplace convenient. Usually the lowest height is defined as zero. The other definitions are: The initial position of the object. The final position of the object. The displacement of the object. Note: If Dy > 0, the object finishes higher than it started. If Dy = 0, the object finishes at the same height as it started. If Dy < 0, the object finishes lower than it started.

9. Other variables: The time the object is in the air. The initial velocity of the object. The final velocity of the object. Note: If any velocity is positive, it means the object is moving upwards. It the velocity is zero, the object is momentarily at rest. If the velocity is negative, the object is moving downwards. All that is left is to substitute these variables into the equations of uniformly accelerated motion.

10. General Equation Freefall Version. V = Vo + gt ∆ y = Vot + 1/2gt2 V2 = Vo2 + 2g∆y ∆y/t = (V +Vo)/2 Again, ‘g’ is the magnitude of the acceleration, and the (-) sign is the direction of the acceleration.

11. These notes are shown with downward as negative. Do this on the overhead projector.

12. Example #1: The south gym measures 9.35 m tall. (a) If an egg is dropped from rest from the top of the gym, how long will it take to reach the bottom? Given information: vo = 0, yo = 9.35 m, y = 0 Dy = y – yo = yo Unknown: t = ? Which equation contains Dy, vo, and t? ∆ y = Vo + 1/2gt2 Substitute Dy = – y and vo = 0

13. Simplify and solve for t: Finally:

14. (b) How fast will the egg be traveling when it hits the bottom? Given information: vo = 0, yo = 9.35 m, y = 0 Dy = y – yo = – yo Unknown: v = ? Which equation contains Dy, vo, and v? Substitute Dy = – yo and vo = 0

15. Square root and substitute numbers: Since the egg is falling downwards, take the negative solution. (Remember, down is negative and up is positive.) Note: You could also solve using the time from part (a):

16. Example #2: How far will an object ideally fall in a time of 10.0 s? How fast will the object be traveling? Assume the object is dropped from rest. Given information: vo = 0, yo = 0 (you can put the origin anywhere) Unknown: v = ? Dy = ? Solution:

17. Example #3: Bubba stumbles upon a deep hole out back of his property. He drops a large stone into the hole, and counts seconds until he hears the sound of the stone hitting bottom. If he counts 16.0 seconds, how deep is the hole? Hint: The stone accelerates downwards from rest. Sound travels back up at a constant velocity of 343 m/s. Trip Downwards: Trip Upwards: Given information: vs = 343 m/s, Distance = h Given information: vo = 0, yo = h, yf = 0 Unknown: h = ? t = time to fall =? (two unknowns!)

18. Solve by setting the two equations for h equal and solve for t. and Putting together:

19. Keep solving! Now solve for h! Don’t fall in!

20. Example #4: Galileo’s Law of Odd Numbers: Galileo studied objects falling to Earth by rolling them down ramps. The ramp allowed to slow the pulling power of Earth’s gravity. Galileo’s results are stated as follows: With each increment of time, a body falls a distance proportional to successive odd numbers. Show that this is equivalent to our uniformly accelerated motion. One unit Four units total Three units Five units Nine units total Sixteen units total Seven units

21. If each consecutive distance follows the prime numbers, then the total distance from the origin is proportional to the square of the falling times. In other words, Use this space to take some notes from the video.

22. Example #5: A ball is thrown upwards at 21.0 m/s from ground level. (a) How high will the ball go? v=0 at top Given information: yo = 0, vo = 21.0 m/s v = vf = 0 at the top. Unknown: yf = h = ? yo=0, vo=21.0 m/s

23. (b) How much time does it take to reach the top? Given information: yo = 0, vo = 21.0 m/s v = vf = 0 at the top. Unknown: t = ?

24. (c) What is the total time in the air? Given information: yo = 0, vo = 21.0 m/s yf = 0 at the bottom. Unknown: t = ? Note: Since the object starts and finishes at the same height, namely ground level, the displacement (Dy) is zero. This has two solutions, one is zero and the other is: This is exactly twice the time to the top.

25. For all freefall motion, the time is symmetric for the trip up and the trip down. An object thrown into the air will take as much time to rise to the top as it will take to come back to the ground. (d) How fast will the ball be traveling when it is 15.0 m above the ground? Given information: yo = 0, vo = 21.0 m/s y = yf = 15.0, Dy = y – yo. Unknown: v = ? The positive solution is for the ball traveling upward towards the top. The negative solution is for the ball traveling back downward to the bottom.

26. (e) How much time does it take for the ball to reach 15.0 m above the ground? Given information: yo = 0, vo = 21.0 m/s y = yf = 15.0, Dy = y – yo. Unknown: t = ? This will be a quadratic, so substitute numbers first and then solve… The only way this equation will be valid is if time is measured in seconds. I will drop the units and just remember the units for t.

27. This equation has solutions: The first solution is for the ball traveling upward towards the top. The second solution is for the ball traveling back downward to the bottom.

28. Example #5: A ball is thrown upwards at 21.0 m/s from the top of the south gym. The gym measures 9.35 m tall. (a) How high will the ball go, as measured from ground level? Given/Unknown information: yo = 9.35 m, vo = 21.0 m/s y = yf = ?, Dy = y – yo. v = vf = 0 at the top.

29. This 22.5 meters is the height of the ball above the top of the gym, not the height from the ground. That height is 22.5 m + 9.35 m = 31.9 m above the ground. (b) How long is the ball in the air? Given/Unknown information: yo = 9.35 m, vo = 21.0 m/s, y = yf = 0 at the bottom, Dy = y – yo. Even though the ball rises up before falling back down, the time in the air can be solved with only one equation. The positive sign on the vo will take care of the upward motion of the ball’s initial motion. This will be a quadratic, so substitute numbers first and then solve…

30. As before, the only way this equation will be valid is if time is measured in seconds. I will drop the units and just remember the units for t.

31. Since the ball cannot hit the ground before you throw it, the negative time is thrown out. Thus the ball hits the ground 4.69 s after it is thrown.

32. (c) How fast is the ball traveling when it hits the ground? Note: The answer is not 0! The answer is the speed just before hitting the ground, not after hitting the ground! Given/Unknown information: yo = 9.35 m, vo = 21.0 m/s, y = yf = 0 at the bottom, Dy = y – yo. Dy = -9.35 m, v = ? Since the ball is traveling downwards at the bottom, the negative solution has been chosen.

33. Note: This could have been solved with a different equation. Since the time was found in the last step, you could also use:

34. Example #6: A model rocket is launched from ground level from rest with an acceleration of +4.00 m/s2. The rocket maintains this constant acceleration for 6.00 seconds, until the fuel is depleted. After this, the rocket moves under the influence of Earth’s gravity. (a) How high will the rocket go, as measured from ground level? This problem must be solved in two parts. The first part is the trip up, with a constant acceleration of + 4.00 m/s2. The second part is the rest of the trip up and back down with an acceleration of –g (freefall). For the first part, two things must be found: The height at the end of the rocket engine’s burn and the speed of the rocket at this point. These will be used later for the second part of the motion. Given information: yo = 0, vo = 0, a = +4.00 m/s2, t = 6.00s Unknown information: y = ?, v = ? (both for the top of the path) First for y:

35. Next solve for v: Now treat the second part as a freefall problem. The initial velocity for the freefall is +24.0 m/s and the initial height for the freefall problem is 72.0 meters. Solve for the max height and the time in the air. Given information: yo = 72.0m, vo = 24.0 m/s, a = - g, vtop = 0 Unknown information: y = ?, v = ? (both for the top of the path)

36. This 29.4 meters is the height above the start of freefall, which itself starts 72.0 m above ground. Thus the total height is:

37. (b) What is the total time the rocket is in the air? During the initial burn of the rocket motor, the rocket flies for 6.00 s. Next we must find how much time the freefall part of the flight takes. Given information: yo = 72.0m, vo = 24.0 m/s, a = - g, ybottom = 0 Unknown information: t = ? Dy = y – yo = -72.0 m As before, t must be measured in seconds, so I will drop the units….

38. Again, disregard the negative solution. The total time in the air becomes the 7.00 seconds of freefall plus the 6.00 seconds of powered flight, for a total of 13.00 seconds.

39. Example #7: An stone falls from a bridge that is 45.0 m above the water. The stone falls directly into a model boat, moving with constant velocity, that was 12.0 m from the point of impact when the stone was released. What is the speed of the boat? Stone Bridge Step 1: Solve for the time to fall to the water if released from rest. Given: vo = 0, yo = 45.0 m, y = 0 Unknown: t = ? 45.0 m drop Boat Solve for the time. 12.0 m distance

40. Step 2: Solve for the speed of the boat.

41. Example #8: A hot air balloon rises at a rate of 12.0 m/s. When the balloon is 80.0 m above the ground, a sandbag is released from rest relative to the balloon. (a) How long does the sandbag take to hit ground? The sandbag is released from rest relative to the balloon, but the balloon is rising at a rate of 12.0 m/s relative to the ground. That means the sandbag is also moving initially 12.0 m/s relative to the ground. Given: vo = +12.0 m/s, yo = 80.0 m, y = 0 Dy = y – yo = - 80.0 m Unknown: t = ?

43. (b) What is the velocity of the sandbag when it hits the ground?