Verticle launch velocity given only time.

[Acadeus]

Homework Statement

you are given the time it takes for a cork to be launched up vertically which is 1.37s

Homework Equations

d=1/2at^2+vt

The Attempt at a Solution

d=1/2(9.8)(1.37)^2
d=9.2m
Because I wanted the height of the parabola I devided both the time and the distance by 2
9.2/2=4.6m, 1.37/2=0.685s
v = 4.6/0.685 = 6.72m/s

 

[tiny-tim]

Hi Acadeus!

you are given the time it takes for a cork to be launched up vertically which is 1.37s
…
Because I wanted the height of the parabola I devided both the time and the distance by 2

I don't understand that

Anyway, why are you finding d?

You have a v and t, and you want u …

use one of the other constant acceleration equations

 

[Simon Bridge]

you are given the time it takes for a cork to be launched up vertically which is 1.37s

Time for what?
Time it takes to launch a cork from first applying the thumbs to when it pops?
OR, what I suspect, time it takes for a cork to reach maximum height?
Presumably neglecting air resistance?

d=1/2(9.8)(1.37)^2
d=9.2m

... so v=0?
What is your reasoning here?

Note: why mess about with d(t) when you know what the final velocity and the acceleration is?

 

[Acadeus]

jeeze i should re-read what I post.
The time i presented is the time the cork took to travel vertically up and down
and I am looking for the launch velocity of the cork.
I used d=1/2at^2 equation so that I could find the full distance that it traveled I just derived it from d=1/2at^2+vt
if that clears anything up

 

[tiny-tim]

jeeze i should re-read what I post.

I used d=1/2at^2 equation so that I could find the full distance that it traveled

the full distance is zero

(alternatively, if the initial velocity is v, what is the final velocity? )

 

[Acadeus]

well I was assuming that the total distance would be 9.2m to travel the full parabola or 4.6m to get to maximum height I was getting there through my own weird means.
I tend to take unorthodox methods to get to answers. I just want to know if it took a cork 1.37s to go up and down if the Launch velocity would be 6.72m/s

 

[Simon Bridge]

the full distance is zero

That would be the total displacement which is zero ;) The total distance traveled would be the length of the trajectory.

well I was assuming that the total distance would be 9.2m to travel the full parabola or 4.6m to get to maximum height I was getting there through my own weird means.

You should use physics instead - works better, and can get much weirder later on.

I tend to take unorthodox methods to get to answers. I just want to know if it took a cork 1.37s to go up and down if the Launch velocity would be 6.72m/s

Well it's those methods that we are here to help with.

To answer your question: there is no reason to believe that you have obtained the correct answer.
i.e. from what you've told us, if you got the correct answer, it's probably a fluke. But I suspect you are close.

There are several ways to approach this. i.e. if it takes 1.37s to go up and down, then how long does it take to just go up to max height? What is the final speed at max height? Which suvat equation includes everything you know except for initial velocity?
OR: if the initial velocity is +u, then what is the final velocity after 1.37s? (same suvat equation)

 

[Acadeus]

well 1.37s is all that I was given in which I am suppose to use to find out the Launch velocity of the cork.

 

[Acadeus]

and 1.37s is the time it took to go up and down

 

[Simon Bridge]

well 1.37s is all that I was given in which I am suppose to use to find out the Launch velocity of the cork.
and 1.37s is the time it took to go up and down

... and that is enough.
You have been given guiding questions in previous responces - why not try answering those questions?

 

[Acadeus]

ok...Thank you for the help I appreciate it.