226 Chapter 2 Projectile Video Questions (20pts)
 
 To receive credit:
  • -Write all answers in your bound journal
  • -2pt if you do not title the entry
  • -2pt if you do not date the entry
  • SHOW ALL YOUR MATH WHEN APPLICABLE

 
Video 223 Parabolic Range (include equations and drawings when asked)
  1. Which projectile angle gives the furthest X-Max?
  2. Which projectile angle gives the highest Y-max?
  3. If the velocity of each projectile is 148m/s, create a table for this video
    like the one you did in the the 222a Parabolic Equations (Velocity - Angle - Vxo - Vyo - 2Tu - Ymax - Xmax)
  4. Explain the equations you set up for lists L1 - L7 and what they determine (Hint 222a).
  5. Draw a diagram showing all the angles and their parabolas shown in the video.
Video 224 Parabolic Vector (include equations and drawings when asked)
  1. Which of the two vector arrows (Vx or Vy) decreases during the video?
  2. Which of the two vector arrows (Vx or Vy) stays constant during the video?
  3. If each dot represents 1 sec, draw the vector diagram at time 0, 7, 11, 14, 21 seconds (they can all be on the same figure).
  4. What force acts constantly on the ball?
  5. Is there acceleration in the Y-direction?
  6. Is there acceleration in the X-direction (remember the definition of acceleration)?
  7. What happens to the Vy vector arrow at 11 seconds? Why?
  8. Under the parabolic graph, there is a cartesian graph for Vx, explain why it remains a constant line over time (t)?
  9. Diagram the cartesian graph for Vx.
  10. Under the parabolic graph, there is also cartesian graph for Vy, explain why the graph starts positive then declines to become zero and finally ends as a negative number?
  11. Diagram the cartesian graph for Vy.
  12. Based on what we talked about in class, Should the absolute value of Vy at time 0s be of the same magnitude as Vy at time 21s?
  13. Why is this the case?
Video 225 Parabolic Soccer (include equations and drawings when asked)
  1. Watch the video and freeze the frame at 30 seconds.
  2. Diagram screen shot and explain what is going on between the X and Y components.
The ultimate bonus question
If you shot a cannon at 45o and the time up and down was 20.0 seconds, what is the X-max of the cannon ball?   to get the bonus question, you must be able to explain it to me and maybe infront of the class
 
2Tu = 20.0 sec so Tu = 10.0 sec Tu = Vy / 9.81m/s^2 Tu * 9.81m/s^2 = Vy 10.0 sec * 9.81m/s^2 = 981m/s = Vy because it is 45o Vy = Vx So therefore Xmax = 2 * Tu * Vx Xmax = 2 * 10.0 sec * 981m/s = 1962 ~ 1960 meters