How long was it in the air before returning to Earth?

[Hydrantman001]

A man jumps to a vertical height of 2.5 m. How long was it in the air before returning to Earth?

here is what i got so far...
vi=0
v=0
x-xi=2.5
a=-9.8

that is for the man jumping upwards...I believe the up and down will be proportional so if I just double my answer in the end i should be alright...

is used the equation x-xi=vi*t + .5((a)t^2)
but I can't seem to get the right answer...
I am thinking that vi is not 0 but instead something else!
Any help would be great!

 

[LeonhardEuler]

A man jumps to a vertical height of 2.5 m. How long was it in the air before returning to Earth?

here is what i got so far...
vi=0

Are you sure $v_i=0$? The man would not move upward if his initial velocity were 0, right? You are using the right equation, though. There is one more tricky part, but I'll let you try to figure it out before I tell you how to do it.

 

[Hydrantman001]

Thanks for replying I figured vi was not 0 but the question does not reveal the true vi to the reader so vi = 7??

 

[LeonhardEuler]

That's the other tricky part. Here's another hint. You know the man's position at two points in time. This gives you two equations.

There is a simpler way of doing it, though. You are right that the upward and downward journeys take the same time. The initial velocity is not 0 for the upward journey, but what about the downward one?

 

[Hydrantman001]

going down the init velocity should be 0 because the object is at rest

 

[LeonhardEuler]

Right, so if you apply the equation there you won't have to worry about v_i.

 

[Hydrantman001]

So 2.5 = -19.6t^2 or -.128 =t^2
so time = .35 or something? is that correct?

and is displacement negative on the way down?
The answer has to be somewhere around 1.4...am i hot or cold?

 

[LeonhardEuler]

If your taking the upward direction as positive, then x-xi=-2.5, not 2.5 because the man starts higher than he ends up on durring the downward trip. Also, its .5a*t^2, not 2a*t^2. And finally, remember that this is only half the journey, so you'll have to double whatever answer you get.

 

[Hydrantman001]

If your taking the upward direction as positive, then x-xi=-2.5, not 2.5 because the man starts higher than he ends up on durring the downward trip. Also, its .5a*t^2, not 2a*t^2. And finally, remember that this is only half the journey, so you'll have to double whatever answer you get.

Yea I am getting about 1.43 which is what I originally got perhaps my webassign is just messed up and won't accept a good answer thanks for the help!

 

[LeonhardEuler]

Yes, that's what I get, too.