Calculating Reaction Time from Dropped Meter Stick

[_physics_noob_]

Homework Statement

A meter stick is held vertically above your hand, with the lower end between your thumb and first finger. On seeing the meter stick released, you grab it with these two fingers. You can calculate your reaction time from the distance the meter stick falls, read directly from the point where your fingers grabbed it. If the measured distance is 17.6cm , what is the reaction time?

So I googled it and it said t=√d/4.9m/s^2
How is that derived?

Homework Equations

v2=v1+at
x=v1t+1/2at^2
x=1/2(v1+v2)t
v2^2=v1^2+2ax

The Attempt at a Solution

I converted 17.6cm to 0.0176m. Now my question is how would I approach this kind of question without all the given variables?
This is my attempt at it.
x=0.176m
a=-9.8m/s^2
0.176=v1t+1/2(-9.8m/s^2)t^2
0=-0.176+v1t+(-4.9m/s^2)t^2
but now I don't have a variable for v1 how to solve for v1

 

[Yenaled]

v1 is the initial velocity. Initially, the meterstick is at rest. Therefore v1 = 0.

 

[Doc Al]

but now I don't have a variable for v1 how to solve for v1

v1 is the initial velocity of the falling object.

 

[_physics_noob_]

So if v1=0 then i could use x=v1t+1/2at^2...
0.176=1/2(-9.8m/s^20)t^2 then solve for t^2
t^2=0.176m/(-4.9m/s^2) but it wouldn't work because I can't square a negative or do i have to use the positive 4.9m/s^2

 

[Yenaled]

You have to make sure your signs are consistent.
Why are you choosing the displacement (x) to be positive but the acceleration (a) to be negative? In this situation, they should be in the same direction (downward).

 

[Rikardus]

Well, the expression really is x - x0 = v0*t + 0.5*g*t^2.

Now the signs will depend on the choice of the x-axis direction. If it points downwards you will assume x = d and x0 = 0 and g = 9.8 (because the gravity pull is in the positive x-axis direction); if you choose an x-axis pointing upwards then x = -d and x0 = 0 and g = -9.8 because the pull is opposite to the positive direction). In each case you get the same answer and there is no negative root to concern yourself with.

 

[_physics_noob_]

Thanks @rikardus ! How did you come up with that expression? did you made
Δx = (v1)t+1/2(a)t^2

 

[Rikardus]

Yes I did. I'm also used to the v0 notation to emphasize that is an initial velocity.

You're welcome.

 

[_physics_noob_]

Okay cool and I figured, that's how it is my book but for some reason my professor uses v1 and v2