Find the height and the velocity of the object

[prishila]

Homework Statement

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.

Homework Equations

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

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?

 

[BvU]

Hi,

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

 

[prishila]

Hi,

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

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?

 

[BvU]

Why the question mark ?

 

[prishila]

Why the question mark ?

I want to know if I'm correct.

 

[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 !

 

[prishila]

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

 

[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.