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For the 2nd part of the question, why can't you use the formula Vf=Vo+2at?

Thanks.

Hi sheumangutman,

Thanks for the question. I think you mean to suggest $v_f = v_o + at$, correct? That approach would be equally fine, provided you establish that $v_f = -v_o$, which is true since it returns back to the original launch height. This means your suggested formula, with a substitution for $v_f$, would become $-v_o = v_o + at$ which rearranges to $t = \dfrac{-2v_o}{a}$ which is the same formula shown at 1:20 in the video. You have to be a little cautious with your suggested formula, however, since it works only when you can tell how the final velocity compares with the initial velocity. $v_f = -v_o$ only when something returns to it's original height. If something fell into a hole after being launched upwards, meaning it returns to a different height, you would be better off using the $d=v_ot + \dfrac{1}{2}at^2$ formula. For this problem, however, the formula you suggest would be just fine.

All the best,
Mr. Dychko

For part c) What specific factors make this an estimate?

Hi Icbishop, did you post this question on the wrong video? I don't see a part c) for this problem.

Cheers,
Mr. Dychko

Well I double checked and I do indeed have a part c on #42 chapter 2. It is the 7th Edition as well. The question also matches up with the solution provided.

Ah, right you are Icbishop. Sorry about that, I just checked the video without looking at the textbook. I've added some notes to the quick answer above the video.

Cheers,
Mr. Dychko

By dividing by "t" instead of factoring aren't we losing a physically meaningful answer t=0 (the time the ball was hit)?

Hi idan, you are making a valid point. You have a sharp eye! Mathematically, it would be more correct to factor and then find the roots of the resulting equation, thereby discovering the answer $t=0$. However, since this question is asking for the "time in the air", $t=0$, as you know, is an extraneous solution, so the effort of finding the solution $t=0$ and then discarding it as extraneous isn't worth the effort. The technique in the video is perfectly fine for a physics class, since math is just a tool for finding solutions to the physical problem, which in this case is "how much time is the ball in the air", not "at what times is the ball at height zero". For the latter, one must follow the technique you suggest by factoring since $t=0$ would be a non-trivial solution, but as it is, $t=0$ is not a solution for determining the "time in the air" problem. In a math class there's no question that to answer "solve this equation", one must factor and include $t=0$ as a solution.

All the best,
Mr. Dychko

Question #39 a ball is thrown straight up with a speed of 36 m/s. how long does it take to return to its starting point. Question is my teacher teaches out of 7th edition I have 6th edition but can not find this question in my book or on this website.It says on top of my page ch# 2 and page ref 2-7 but cant find it also how come you don't post question you just go to answers only . need to see question sometimes so I can find which edition it is in.

Hi EddieG, thanks for the question. I would love to post the questions, but I've avoided that since it would be a copyright issue with the publisher since I didn't create the questions. Perhaps you can ask your teacher to photocopy just the problems from their 7th Edition text for your reference?
All the best,
Mr. Dychko