# Math Help - Free Fall Lmit/Derivative problem

1. ## Free Fall Lmit/Derivative problem

An object is dropped from the top of a 100-m tower. Its height above the ground after 1 sec is 100-4.9t^2 m. How fast is it falling 2 sec after it is dropped?

Formula:
lim as h--> 0 (f(t+h)- f(t))/h
or as an easier formula, I think.
lim as h--> 0(f(t) - f(h))/(t-h)

Guess: 100=4.9t^2 m goes as t and solve. Then plug in 2 for t.

Thank you!

2. Originally Posted by Truthbetold
An object is dropped from the top of a 100-m tower. Its height above the ground after 1 sec is 100-4.9t^2 m. How fast is it falling 2 sec after it is dropped?

Formula:
lim as h--> 0 (f(t+h)- f(t))/h
or as an easier formula, I think.
lim as h--> 0(f(t) - f(h))/(t-h)

Guess: 100=4.9t^2 m goes as t and solve. Then plug in 2 for t.

Thank you!
the problem says the formula is for the height after 1 seconds. doesn't that mean we plug in t = 1 for 2 seconds, since it is one more second after the formula applies? i don't know, that's oddly worded. anyway, what you want is:

$\lim_{h \to 0} \frac {f(1 + h) - f(1)}{h}$

i used 1 here for 1 additional second. maybe you should use 2 instead.

3. Ah. Syntaz LaTex error.
*you are seeing Jhevon's formula quoted*

How do I get rid of the h on the bottom? My book peruses by it without telling.
Something about 1/h times the whole thing.
The answer in the book is: 19.6 m/sec.
After I figure out the way above equation, do I put $100-4.9t^2$ as a and solve?
Thanks.

4. Originally Posted by Truthbetold
Ah. Syntaz LaTex error.
*you are seeing Jhevon's formula quoted*

How do I get rid of the h on the bottom? My book peruses by it without telling.
Something about 1/h times the whole thing.
The answer in the book is: 19.6 m/sec.
After I figure out the way above equation, do I put $100-4.9t^2$ as a and solve?
Thanks.
ok. so we should plug in t = 2 then. fine.

(did you try to work it out? it seems pretty straight forward to me)

$\lim_{h \to 0} \frac {f(2 + h) - f(2)}h$

$= \lim_{h \to 0} \frac {100 - 4.9(2 + h)^2 - 80.4}h$

$= \lim_{h \to 0} \frac {19.6 - 4.9 \left(4 + 4h + h^2 \right)}h$

$= \lim_{h \to 0} \frac {-19.6h - 4.9h^2}h$

$= \lim_{h \to 0} (-19.6 - 4.9h)$

$= -19.6$

(the velocity is negative because we are falling, that is, moving in the negative direction)

5. I got it now.
It was all because in - f(a), why I'm being clueless and asking odd question.
I would write the equation as - 100 -4.9(2)^2.

there we go. The answer to half the problems!

Thanks you. Yes, thanks you.

6. I spoke too soon.

You lost me here.

how did you do that?
It seems you stuck 4h with h as 0 to cancel it out. After that, I don't know what you did.
Thank you.

7. Originally Posted by Truthbetold

I spoke too soon.

You lost me here.

how did you do that?
It seems you stuck 4h with h as 0 to cancel it out. After that, I don't know what you did.
Thank you.
fine, i'll show all the steps. i assume you understand everything up to that point.

$\lim_{h \to 0} \frac {19.6 - 4.9 \left( 4 + 4h + h^2 \right)}h$ ..........expand the brackets

$= \lim_{h \to 0} \frac {19.6 - 19.6 - 19.6h - 4.9h^2}h$ ........simplify

$= \lim_{h \to 0} \frac {-19.6h - 4.9h^2}h$ ...........factor out the h

$= \lim_{h \to 0} \frac {h(-19.6 - 4.9h)}h$ ............cancel the h

$= \lim_{h \to 0} (-19.6 - 4.9h)$ ..............take the limit

$= -19.6$