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.
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.
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?
Put in the time it hits the ground!?? You want to know how fast it's going 'when it hits the ground'. The time you just figured out. You do know that v(t) is the velocity at time=t, right?
I want to find the velocity the instant before it hits the ground. I guess I'm misunderstanding the question and I should be finding it for t = 8√2/√g and not trying to figure out how to express a value for time that's an instant before. :grumpy: 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?
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.
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.