Free Fall and other Animals

HANDOUT PROBLEMS:
4. 706.6 meters
5. 4.9 meters
6. 400 meters
7. 1.078 seconds
8. 7.877 m/s

HW SOLUTIONS: Page 43,Problems 34, 37, 38, 41, 44, 46

34. 60.03 meters

37. From the max height, you can get the initial velocity of the kangaroo. (7.275 m/s) You can then use this with a kinematics equation to get the time required to reach this height. Double it to get the answer of 1.485 s.

38. Since the total time is 3.3 seconds, it takes 1.65 seconds to reach the top. You can use this to find the initial velocity of the ball (16.17 m/s) from v = v0 + a*delta_t. The height can then be found using v^2 = vo^2 + 2a*delta_x. The answer is 13.06 meters.

41. Set up a quadratic with delta_y = -105 meters, v_o = 5.5 m/s, and a = -9.8 m/s². Solving numerically, delta_t = 5.224 seconds.

44. You can use v^2 = vo^2 + 2a*delta_x to find the initial velocity　of 12.837 m/s. Again, set up a quadratic to find the times. There are two times because the object has the 12 meter height both when it goes up and goes down. These times are at 0.7309 s and 3.351 seconds.

46. Follow the hint given on the sheet - first try to find the initial velocity of the stone at the top of the window. Since acceleration is constant, we can use the equation stating that average velocity = (v + v0)/2 = delta_x/delta_t. This means the velocity of the stone halfway through its fall past the window is 7.333 m/s. This means that after 0.15 s (halfway of the fall time), the speed of the stone is 7.333 m/s.

For the interval of time from when the stone passes the top of the window to when its speed is 7.333 m/s, v = 7.333, delta_t = 0.15 s, a = 9.8 m/s² downwards, and v0 is unknown. Calculating, v0 = 5.863 m/s.

Now consider the interval of time between when the stone is dropped and it FIRST reaches the top of the window. During this interval, v0 = 0 m/s, a = 9.8 m/s², and v = 5.863 m/s. From these quantities, you can use ONE of the kinematics equations to find the displacement of the stone over this interval. The final answer is 1.754 meters.