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

 

[Nickie]

I would approach this problem by first identifying the given information and what is being asked. The problem provides the height of the window, the time it takes for the rock to pass the window, and the acceleration due to gravity. The questions then ask for the velocity at the top and bottom of the window, the time it takes for the rock to reach the top of the window, and the initial height of the rock.

To solve for the velocity at the top of the window, we can use the equation v = vo + at, where v is the final velocity, vo is the initial velocity (which is 0 in this case), a is the acceleration due to gravity, and t is the time it takes for the rock to reach the top of the window. Substituting the given values, we get v = 0 + (-9.81m/s^2)(0.164s) = -1.61m/s. Since the rock is moving downwards, the velocity at the top of the window is negative.

To solve for the velocity at the bottom of the window, we can use the same equation, but this time the time is the total time it takes for the rock to reach the bottom of the window, which is twice the time it took to reach the top. Thus, t = 2(0.164s) = 0.328s. Substituting this value, we get v = 0 + (-9.81m/s^2)(0.328s) = -3.22m/s.

To find the time it takes for the rock to reach the top of the window, we can use the equation h = vot + 1/2at^2, where h is the height of the window, vo is the initial velocity, and t is the time it takes to reach the top of the window. Substituting the given values, we get 2.00m = (0)(t) + 1/2(-9.81m/s^2)(t^2). Solving for t, we get t = 0.164s, which is the same as the given time.

To find the initial height of the rock, we can use the same equation, but this time solving for h. Thus, h = 1/2at^2 = 1/2(-9.81m/s^2)(0.164s)^2