# Homework Help: Find the height and the velocity of the object

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1. Nov 26, 2015

### prishila

1. The problem statement, all variables and given/known data
A bolt is detached from the lower part of an elevator cabin that is ascending with velocity 6meters/second and for 3 seconds comes to the point from where the elevator started to ascend. Find:
in which height is the end of the elevator in the moment when from it the bolt was detached. What is the velocity of the bolt when it comes to the point the elevator started to ascend.

2. Relevant equations
h=v0t+(gt^2)/2
v^2-v0^2=2gs

3. The attempt at a solution
Here's what i did:h=v0t+(gt^2)/2 vo is the first velocity
h=18+(10*9)/2=63
v^2-v0^2=2gs
v^2=2gs+vo^2=2*10*63+36 v=36
But my teacher told me that I did it wrong, because the bolt firstly goes up (maybe after it is detached from the elevator) and then goes down. Could you correct my mistake?

2. Nov 26, 2015

### BvU

Hi,

In "h=v0t+(gt^2)/2" v0 goes up and g goes down. Yet you give both the same sign

3. Nov 26, 2015

### prishila

So I should put v0t-gt^2/2=-27, so the end of the elevator is 27 meters above the point the elevator was when the bolt departed?

4. Nov 26, 2015

### BvU

Why the question mark ?

5. Nov 26, 2015

### prishila

I want to know if I'm correct.

6. Nov 26, 2015

### BvU

That's for teacher to decide ! But I think you can be a lot more confident with this value than with the original 63 m !

7. Nov 26, 2015

### prishila

And then I can use the formula v^2-vo^2=2gh?

8. Nov 26, 2015

### BvU

Depends on what h means here. Would you substitute the -27 you found for h or the +27 m you answered as "end of the elevator is 27 meters above the point the elevator was when the bolt departed" ?

In fact your expression $$h = v_0 t + {1\over 2} a t^2$$ (with a = -10 m/s2) is better formulated as $$h = h_0 + v_0 t + {1\over 2} a t^2$$ so that you calculate $h - h_0 = -27$ m .
$h$ is where the bolt is after 3 sec: at the starting point of the elevator. We can call that h = 0.
$h_0$ is where the bolt is at t=0 when it comes apart from the elevator.
This way h0 comes out positive. And ${1\over 2} mv^2 - {1\over 2} m v _0^2 = mgh_0$ indeed.

Another, easier equation is $v = v_0 + at$ with v0 = 6 m/s and a = -10 m/s2. It also directly provides the correct sign.