+2440 Hi Melody,
Ok . . . I did not know that. I was really scratching my chimp head on this.
I did note that she was totally lacking in gratitude with words or points, but such behaviors are not rare on here.
By dumb luck, I managed to snag copies of a few of the better question posts shortly after the answers posted, but before she deleted the questions. I noticed two deleted posts; I assumed they were malfunctions. I should have been aware though, because she did the same thing with her Sapchats, leaving 40+ followers without a history, except for a statement about being too lazy to post. It’s not true; no one has that many followers with no posts.
Her aberrant behavior seems to be integral to her current personality. If she does return, it’s likely her behavior will too. One of my psychology professors is always interested in anecdotal observations of anomalous behavior. I may relay this to him in the fall.
Two years ago, Nauseated noticed a forum member who started deleting his older posts—his answers to questions. He deleted six pages worth, and then quit doing it after posting the reason why. You may remember this: http://web2.0calc.com/questions/good-luck-with-this-one#r11
Naus said, “It’s only a few picoteslas. It’s not a great loss. I’m not sure if he was serious or being his trolling self—probably both. He did add if Heureka, Alan, CPhill, Rom, Bretie, or Melody did it, then it would be like having a major fire in the mathematics section of a unique library.
In the case of Miranda, I think she should have pressed ‘Ctrl-Alt’ before pressing delete. Her questions are not unique, but the solution presentations are, and those questions were the signposts for the pathways.
Maybe we can restore a few of these. I can post the questions as a guest and the original answerers can copy their answers to the new post.
GA
P.S. Have a nice day
.
+2440 Solutions for A - E
\(\text{Position and Time }\\ \begin{array}{|rcll|} \hline a_1) \text { position }\\ h &=& v_i(t) -\dfrac{1}{2} g(t)^2 \\ h &=& 15t - 4.9t^2 \\ t\tiny \; \text{ @} \small \text { 1 and 4}\\ h &=& 15(1) - 4.9(1)^2 &=&10.1m \\ h &=& 15(4) - 4.9(4)^2 &=&-18.4m \\ a_2) \text { velocity} \\ v_f &=& v_i – gt \\ v_f &=& 15 - 9.8t \\ \text{t @ 1 and 4} \\ v_f &=& 15 - 9.8(1) &=&5.20m/s \\ v_f &=& 15 - 9.8(4) &=& -24.2m/s \\ \hline \end{array}\\ \text { at 1 second: ball is 10.1m (above) and moving @ 5.20 m/s (up)} \\ \text { at 4 seconds: ball is -18.4m (below) and moving -24.2 m/s(down)} \\\)
\(b)\text {Gravity is slowing ball} \\ \begin{array}{|rcll|} \hline v^2 &=&v_0^2 +2(a)(\Delta y) \\ v^2 &=&(15)^2 +2(-9.8)(5m) \leftarrow \text{ use negative acceleration} \\ v &=&\pm \sqrt{127} \text{ (use positive result) } \\ v &=& 11.27m/s \\ \hline \end{array}\)
\(c)\text {Maximum height occurs when the vertical velocity is zero (0)} \\ \begin{array}{|rcll|} \hline a &=& \dfrac {(v_f – v_i)}{t} \\ t &=& \dfrac {(v_f – v_i)}{a} \\ t &=& \dfrac {(0 - 15)}{(- 9.8)} \\ t &=& \dfrac{-15}{-9.8} \\ t &=& 1.53s \\ \hline \end{array}\)
\(d) \text{ Acceleration is constant (g)} \\\)
\(e) \text{ Time after displacement }\\ \begin{array}{|rcll|} \hline t &=& \dfrac {(v_f – v_i)}{a} \\ v_f^2 &=& v_i^2 + 2(a)(\Delta y)\\ v_f &=& \pm \ sqrt{(v_i^2 + 2(a)(\Delta y)} \leftarrow \small \text {Use negative result; the ball is moving downward } \\ t &=& \dfrac{-\sqrt{(v_i^2 + 2(a)(\Delta y)} + (v_i)}{(-g)} \\ t &=& \dfrac{-\sqrt{(15^2 + 19.6*5)} + 15)}{-9.8}\\ t &=& 3.36s \\ \hline \end{array}\)
For related effects of gravity, see this.
