Solving the Rock Falling: Velocity & Time

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Homework Help Overview

The problem involves a rock being dropped from a building and passing a window of known height, with the goal of determining the velocities at the top and bottom of the window, the time taken to reach the top, and the height from which it was dropped. The subject area is kinematics, specifically dealing with motion under constant acceleration due to gravity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply kinematic equations but expresses uncertainty about their application and the results. Some participants suggest using different equations to clarify the relationships between the variables involved.

Discussion Status

The discussion is ongoing, with participants exploring various equations and approaches to the problem. There is a recognition of confusion regarding the correct application of kinematic equations, and some guidance has been offered to redirect the focus toward equations that involve a single unknown.

Contextual Notes

Participants are navigating the constraints of the problem, including the need to account for multiple unknowns in the equations being used. There is also a mention of the specific time interval during which the rock passes the window, which is critical to the calculations.

Reth0407
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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.
 
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Welcome to PF!

Hi Reth0407! Welcome to PF! :smile:

Try s = ut + 1/2 at2 :wink:​
 
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?
 
6.26 ? :confused:

oh, that's from …​
Reth0407 said:
2a(x-xo)=vx2-vox2
2a(x)=-vox2
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 - xo = vot + 1/2 at2
 

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