Your 'cq_1_8.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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cq_1_82
Your 'cq_1_8.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).
· How high does it rise and how long does it take to get to its highest point? (15m/s)^2 / 2a = (15m/s)^2 / 2(-10m/s/s) = 11.25m + 12m above the ground = 23.25m above the ground
· How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground? Don’t know
You know v0, a and `ds:
Assuming upward to be the positive direction, v0 = 15 m/s and the acceleration (being downward, in the direction opposite to the chosen positive direction) is a = -10 m/s^2.
Between the initial instant and striking the ground the ball has a displacement of 12 meters downward. This is the displacement from the initial position to the ground. So `ds = -12 m.
Which equations of motion apply when you know v0, a and `ds?
&&&& vf^2 = vo^2 + 2a’ds. Vf^2 = (15m/s)^2 + 2(-10m/s/s)(-12M) = 225(m/s)^2 + 240(m/s)^2 = vf = 21.56m/s
The final velocity will be in the downward direction. Solving vf^2 = 465 m^2/s^2 for vf, you get vf = +- 21.56 m/s. You have to consider both the + and the -. In this case the + solution would be discarded.
&&&& vf = vo+a*’dt ; vf – v0 = a’dt ; (vf-v0)/a = ‘dt
&&&& (21.56m/s – 15m/s) / -10m/s/s = ‘dt ; 6.56m/s / -10m/s/s = .66s That doesn’t look right.
You are right. It wouldn't be that quick. Good attention to detail.
Note also that your calculation would acutally end up as -.66 s.
If you use the - solution for vf you will get the correct answer.
· At what clock time(s) will the speed of the ball be 5 meters / second? Don’t know
Assuming upward to be the positive direction, v0 = 15 m/s and the acceleration (being downward, in the direction opposite to the chosen positive direction) is a = -10 m/s^2.
The relevant interval is the interval between the initial instant and the instant the ball's velocity reaches 5 m/s. On this interval the final speed of the ball is 5 m/s, so its velocity is either 5 m/s or -5 m/s.
In either case you know v0, vf and a.
How do you reason out a situation in which you know v0, vf and a?
&&&& We can find clock time by ‘dv / a = 10m/s / -10m/s/s = 1s
`dv = vf - v0 = -5 m/s - 5 m/s = -10 m/s, which is a good thing because 10 m/s / (-10 m/s/s) = -1 sec, not 1 sec.
How do the equations of uniformly accelerated motion apply to a situation where you know v0, vf and a? vf = v0+a*’dt
Now you have two choices for vf. Consider each choice separately:
Assuming vf = 5 m/s, you can find `dt and `ds. What do you get?
&&&& ‘ds = 1s; ‘ds = vAve*t = 10m/s * 1s = 10m
Assuming vf = -5 m/s, you can find `dt and `ds. What do you get?
&&&& ‘dt = vAve/a = 5m/s / -10m/s/s = .5s
&&&& ‘ds = vAve*t = 5m/s * .5s = 2.5m
· At what clock time(s) will the ball be 20 meters above the ground? Don’t know
Assuming upward to be the positive direction, v0 = 15 m/s and the acceleration (being downward, in the direction opposite to the chosen positive direction) is a = -10 m/s^2.
For the interval of interest here, the ball will displace +8 m from the initial point to its highest point. So `ds = 8 m.
So you know v0, a and `ds. How do you solve a situation in which you know v0, a and `ds?
&&&& vf^2 = vo^2 + 2a’ds ; vf^2 = (15m/s)^2 + 2(-10m/s/s)(8m) = 225(m/s)^2-160(m/s)^2 = 65 (m/s)^2; vf = 8.06 m/s. I believe that was useless for this question, but did I find vf correctly?
What do you get in this case?
&&&& Since I already found vf, I can use vf = v0 + a*’dt ; ‘dt = (vf-vo) / a = (8.06m/s – 15m/s) / -10m/s/s = -6.94m/s / -10m/s/s = .69s That doesn’t look right either!
In this situation the negative solution for vf would correspond to the situation.
· How high will it be at the end of the sixth second? Don’t know
Assuming upward to be the positive direction, v0 = 15 m/s and the acceleration (being downward, in the direction opposite to the chosen positive direction) is a = -10 m/s^2.
For this interval `dt = 6 seconds.
So you know v0, a and `dt. How do you solve a situation in which you know v0, a and `dt?
What do you get in this case?
&&&& Will the final velocity after 6 sec be –45m/s/s? Change in position can be found by vAve * ‘dt = -15m/s * 6s = 90m
-15 m/s * 6 s = -90 m; the displacement would be -90 m (if acceleration remains uniform for 6 second it's probably because there's a hole in the ground).
answer/question/discussion:
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i spent about 1 hr on this seed question and the previous one.
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I'm lost. I tried the book but I didn't get far.
For each of the above situations you know v0 and a. Different situations imply different intervals, so in some cases you know vf, in some you know `dt, in some you know vf. Putting any of these three together with v0 and a you have enough information to either solve the equations for or reason out all other quantities.
See my notes.
Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&&.
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I'm still struggling for some reason. I'm getting answers that seem to be inconsistent.
You've almost got it. See my notes and send another revision; this time use #### to indicate insertions into this document. You've done almost all the calculations correctly so this shouldn't take you long.