How High Was the Elevator When the Nut Fell?

[SS2006]

A nut falls off of an elevator moving upwards at 3 m/s. The nut takes 5 s to hit the ground.

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

 

[Ouabache]

Ooops... Well on the bright side, I find solving a problem more than once, reinforces the concepts I'm learning..

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.

 

[SS2006]

well i just can't reach 107.5 m
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

 

[iNCREDiBLE]

v = v_0 + at
&
v^2 - v_0 = 2as

 

[Integral]

This is pretty straight forward,
simply plug your time (5s) into

d = - \frac {gt^2} 2 + V_0t + h

What would your V_0 and h be?

 

[Ouabache]

Did you draw a diagram of your problem, with arrows indicating which direction velocity is moving, which direction acceleration is moving and positions of object?

Let's see, what does the first equation (v2=v1+at) tell you?
Given an intial velocity v1, and an acceleration a, it tells you an object's velocity after some time t.

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 direction in your equation (up or right as + , down or left as -). Using this concept, make sure you use the appropriate signs + or - , in your 2nd equation.