
#1
Dec308, 09:33 PM

P: 9

1. The problem statement, all variables and given/known data
When Mongol drops his ball off the Hasselbluff Mountain Viewpoint, its height above the ground at time t is given by h = 64  1/2gt^{2}. How long will the ball take to hit the ground? How fast is the ball going the instant before it hits the ground? 2. Relevant equations I found the answer to the first question to be: t = 8√2/√g The derivative is v(t) =  gt 3. The attempt at a solution v(t) = [lim t>0]  g(8√2/√g + ∆t)   g(8√2/√g) v(t) = [lim t>0]  g8√2/√g + g(8√2/√g) v(t) = [lim t>0] 0 I know that I'm doing something very obviously wrong, I just can't figure out what. 



#2
Dec308, 10:06 PM

P: 179

You already found v(t), so why not just evaluate [tex]v \left( \frac{8\sqrt{2}}{\sqrt{g}} \right)[/tex]?
It looks like you are trying to find acceleration with that limit, but incorrectly. 



#3
Dec308, 10:06 PM

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P: 25,171

You said v(t)=gt. Very true. Why don't you just use that? What's this limit stuff all about?




#4
Dec308, 11:26 PM

P: 9

Finding velocity of a falling object
Well my reasoning was that since it hit the ground at t = 8√2/√g. I would have to put a limit on t to make it very, very close to that but not equal to it since then it would have already hit the ground. I suppose for what I was going for, the limit would have been 8√2/√g.
So what time would I insert into the equation v(t)=gt? 



#5
Dec408, 07:54 AM

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P: 25,171





#6
Dec408, 09:56 PM

P: 9

So, plugging it into the equation I get: v(t) =  g(8√2/√g) v(t) =  8g√2/√g v(t) =  8(9.8)√2/√(9.8) v(t) ~ 35.42 ...which is definitely the correct answer. Thank you so much Dick and mutton! My misunderstanding definitely came from thinking that if I put in the time at which it hit the ground, the result for velocity would be zero. Why doesn't this happen? 



#7
Dec408, 11:19 PM

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That's a question for philosophers. When it is "on" the ground it's velocity is zero. An instant before it hits the ground, it's exactly what you computed. That's what they are after.




#8
Dec408, 11:31 PM

P: 179

It's implied that the given function h is valid only when t [tex]\ge[/tex] 0 and h [tex]\ge[/tex] 0. Notice that h < 0 for t > 8√2/√g, which would mean the ball goes underground.
So what you found is a limit; it is the limit of v as t approaches 8√2/√g, because v is really 0 when t = 8√2/√g. "The instant before it hits the ground" isn't just any instant; this wording lets you describe what happens if the ball were to keep falling. 



#9
Dec608, 09:38 AM

P: 9

Oh wow! Got it! Thanks again you two, you've been a great help.



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