march.26th.2009

At the beginning of class today we had to answer the following question: Why does the vertical component of velocity for a projectile change with time, where as the horizontal component does not? The answer is the vertical component has forces acting on it while the horizontal component does not have any forces acting on it. Mr. Manning then asked for Hershey Park money that is due tomorrow. He explained that if we decide not to go that we still have to do the packet only with fake numbers and not the actual Hershey Park numbers. After we talked about the Hershey Park money we started to work on our Projectile Trajectory Activity which is page 6 in our notebooks. We started this worksheet at the end of class yesterday and today we answered question 3a and 3b. The answers were: 3a. -9.8 m/s squared and 3b. 0 m/s squared. He then asked us to work on the Horizontal Motion chart on the worksheet. To find the answers for the chart all you have to do is multiply the time by 20 and it will give you the delta x in meters. After we worked on the Horizontal Motion chart we had to work on the Vertical Motion chart. The charts with the answers are copied below. Horizontal Motion chart Vertical Motion Chart We then worked on the graph on the back of the worksheet which was page 7. After we finished the graph we worked on a problem on the smartboard. It involved finding the initial speed at which a boy threw a ball from a tower 4.9 meters high and the ball was thrown 20 meters. The time was 1 second. The velocity(answer) was 20 m/s. We were given page 8 and 9 to do at the end of class and if we don't work it will end up being homework.
 * = T(s) ||= ∆x(m) ||
 * = 0 ||= 0 ||
 * = 1 ||= 20 ||
 * = 3 ||= 60 ||
 * = 5 ||= 100 ||
 * = 7 ||= 140 ||
 * = 9 ||= 180 ||
 * = T(s) ||= ∆x(m) ||
 * = 0 ||= 0 ||
 * = 1 ||= 4.9 ||
 * = 3 ||= 44.1 ||
 * = 5 ||= 122.5 ||
 * = 7 ||= 240.1 ||
 * = 9 ||= 396.9 ||