Solving the Rock Falling: Velocity & Time

[Reth0407]

Homework Statement

A rock is dropped off the top of a building. On the way down, the rock passes a window. The window is known to be 2.00m tall and the stone takes 0.164s to fall past the window.

1)What is the velocity of the rock at the top of the window?
2)What is the velocity of the rock at the bottom of the window?
3)How much time was necessary from the instant the rock was dropped until it reached the top of the window?
4)How far above the top of the window was the rock dropped?

I plugged this but wasn't sure if I started it right.
xo=0 vox= 0 ax=-9.81m/s^2 t=0.164s x=2.00m Vx=?

Homework Equations

2a(x-xo)=vx^2-vox^2
2a(x)=-vox^2
sqrt 2(9.81m/s^2)(2.00m) = 6.26 m/s? not sure if this is the answer for 1). But I pretty sure it can't be right since question 1 and 2 is asking for top velocity and bottom velocity, which is why I'm confused on how to solve this.

The Attempt at a Solution

Attempted once but got stuck on how to continue solving the rest.

 

[tiny-tim]

Welcome to PF!

Hi Reth0407! Welcome to PF!

Try s = ut + 1/2 at² ​

 

[Reth0407]

Hi tiny-tim! Ok that equation would make sense for half. So velocity of rock at bottom would be the full right which is 6.26m/s?

 

[tiny-tim]

6.26 ?

oh, that's from …​

2a(x-x_o)=v_x²-v_ox²
2a(x)=-v_ox²
sqrt 2(9.81m/s^2)(2.00m) = 6.26 m/s?

no, both those vs are non-zero, why have you left one out?

this is the wrong equation (it doesn't help, because you have two unknowns in it)

you need a constant acceleration equation with only one unknown in it, ie x - x_o = v_ot + 1/2 at²