- #1

SS2006

- 80

- 0

Q) How high was the elevator the instant the nut fell off , the answer should be 107.5 metres

i solved this a few weeks ago, now i just can't remmeber it :( casue the test is in 2 days

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- Thread starter SS2006
- Start date

- #1

SS2006

- 80

- 0

Q) How high was the elevator the instant the nut fell off , the answer should be 107.5 metres

i solved this a few weeks ago, now i just can't remmeber it :( casue the test is in 2 days

- #2

Ouabache

Science Advisor

Homework Helper

- 1,343

- 8

How about posting what you have tried so far and show where you are getting stuck? There are many here, who can help steer you in the right direction.

- #3

SS2006

- 80

- 0

ive tried v2=v1+at

and d = v1t + 1/2 at squared, the ones that worked last time, i think I am using the wrong numbers

- #4

iNCREDiBLE

- 128

- 0

[tex]v = v_0 + at[/tex]

&

[tex]v^2 - v_0 = 2as[/tex]

&

[tex]v^2 - v_0 = 2as[/tex]

- #5

Integral

Staff Emeritus

Science Advisor

Gold Member

- 7,253

- 63

simply plug your time (5s) into

[tex] d = - \frac {gt^2} 2 + V_0t + h[/tex]

What would your [itex]V_0[/itex] and h be?

- #6

Ouabache

Science Advisor

Homework Helper

- 1,343

- 8

Let's see, what does the first equation (

Given an intial velocity

But you are interested in height (or distance), so first equation doesn't do very much for you.

If you drew arrows indicating the direction of your variables on your initial diagram, you want to also indicate

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