Your 'cq_1_8.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
Assignment 8Seed Question 1
A ball is tossed upward with an initial velocity of 25 meters / second. Assume that the acceleration of gravity is 10 m/s^2 downward.
• What will be the velocity of the ball after one second?
answer/question/discussion: At one second, the velocity of the ball should be 35 m/s. The reason I think this is because a = ‘dv / ‘dt. If a = 10m/s/s and v0= 25 m/sec and the time change is one sec, then we can find vf by multiplying 10 m/s/s by 1 sec and then adding 25 m/sec to get 35 m/sec.
Good reasoning, but the (downward) acceleration of the ball is in the direction opposite its (upward) velocity. Its velocity after 1 second is therefore 15 m/s.
What will be its velocity at the end of two seconds?
answer/question/discussion: The velocity at the end of two seconds should be solved in the same manner, but ‘dt as the denominator will be 2 sec instead of 1. Thus, the acceleration will be 45 m/s/s.
• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: Therefore, during the first two seconds the average velocity would be vf of 45 m/sec + v0 of 25 m/sec divided collectively by two to get 35 m/s.
• How far does it therefore rise in the first two seconds?
answer/question/discussion: Therefore, it rises 70 m, determined by the ‘ds equation ‘ds=vAve+’dt and subbing in 35 m/s for vAve and 2 sec for ‘dt, which gives us the change in position.
Your analysis is faultless, except that the acceleration is in the downward direction.
You have chosen to make the upward velocities positive, so you need to make any downward quantity negative. Your acceleration should therefore be -10 m/s^2.
• What will be its velocity at the end of an additional second, and at the end of one more additional second?
answer/question/discussion: At the end of three seconds, the velocity would be found using the same reasoning as before and would be equal to 55 m/s and then at four seconds it would be 65 m/s.
• At what instant does the ball reach its maximum height, and how high has it risen by that instant?
From this point on, the difference in signs becomes even more essential. Your reasoning in these questions doesn't always hold up, but if you redo the previous questions and think about what it happening, you should then be able to come to the correct conclusions here.
answer/question/discussion: The instant that the ball reaches its maximum height is when vf takes over as 0 m/s, which would happen at about 25 m, since if vf is set equal to 0, the v0 is still 25, and the overall velocity change would be -25 m/s. At this point, there have been 25 m covered.
• What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?
answer/question/discussion: The vf in the first four seconds is equal to 65 m/s, as noted above. Therefore, we can find the vAve by taking (65m/sec-25m/s) / 2 = 20m/s. The height at the end of this time interval is 20m/s * 4s = 80m.
• How high will it be at the end of the sixth second?
answer/question/discussion: The height at the end of the 6th second will be solved in the same manner as above; we find vAve to be (85m/s-25m/s) / 2 = 30m/s. Then we multiply to get our height, as 30m/s * 6s = 180 m/s.
** **
It took me about 25 minutes to do this assignment.
** **
Very good reasoning. You do have an error in the relative signs of your velocity and acceleration. See my notes and please revise accordingly. You should not find it diffucult or time-consuming to do so--most of your explanations can simply be copied word for word, with changes only in the numbers.
Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&&.