What is the final velocity of a stone thrown downwards from a roof?

  • Thread starter Reth0407
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In summary, a stone is thrown vertically downwards from a building with an initial speed of 16.0m/s and a height of 58.2m. Using the equation 2a(x-xo)=vx^2-vox^2, the time of fall is found to be 3.88 seconds. The final velocity of the stone can then be calculated by rearranging the equation and solving for vx, resulting in a final velocity of 54.1m/s.
  • #1
Reth0407
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Homework Statement


A person throws a stone vertically downward with an initial speed of 16.0m/s from the roof of a building 58.2m above the ground.
What velocity does the stone have at impact?


Homework Equations



initial x= 58.2m initial vox=16.0m/s ax= 9.81m/s^2 xfinal=0 vx=?

2a(x-xo)=vx^2-vox^2 to find t= 3.88

vx=vox+axt plug in a number including t=3.88 vx= 54.1



The Attempt at a Solution


2 attempts with only 2 attempts left. Please help
 
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  • #2
Simplest way is use nergy conservation: initial kinetic energy + loss of potential energy = final kinetic energy.

Other way: v = v0+ gt
h = ∫v(t)dt = v0t + gt2/2 & solve for t = time of fall
h = height of building
Then stick t into the v equation above.
 
  • #3
I am still confused on how I would plug the numbers because the instructor have me equations with v and x variables. However, are my numbers correct for time=3.88 s and my vx = 54.1 m/s which I'm sure isn't correct. I'm solving for vx
 
  • #4
Reth0407 said:

Homework Statement


A person throws a stone vertically downward with an initial speed of 16.0m/s from the roof of a building 58.2m above the ground.
What velocity does the stone have at impact?


Homework Equations



initial x= 58.2m initial vox=16.0m/s ax= 9.81m/s^2 xfinal=0 vx=?

2a(x-xo)=vx^2-vox^2 to find t= 3.88
t doesn't appear in the equation, so how do you obtain t from it :confused: (Also, that doesn't look like a correct value for t for this situation).

However, you can use that equation to find the final velocity, vx. Rearrange the equation to solve for vx.
 
  • #5
Ok I got it! LIke gneill said, I could rearrange the equation and solve for vx. Thank you all.
 

1. What is the "A Stone Thrown Downward" experiment about?

The "A Stone Thrown Downward" experiment is a classic physics experiment that investigates the motion of a falling object. It involves dropping a stone from a certain height and recording its position and velocity at different points in time.

2. What is the purpose of the experiment?

The purpose of the experiment is to study the effects of gravity on a falling object and to determine the relationship between an object's position, velocity, and time during free fall.

3. How is the experiment performed?

The experiment is performed by dropping a stone from a known height and recording its position at regular intervals using a measuring device such as a ruler or stopwatch. The stone is dropped in a vacuum to eliminate air resistance and ensure more accurate results.

4. What are the key variables in the experiment?

The key variables in the experiment are the stone's initial height, its position at different points in time, and its velocity at those points. These variables are measured and recorded to analyze the stone's motion and determine the acceleration due to gravity.

5. What are some potential sources of error in the experiment?

Some potential sources of error in the experiment include air resistance, human error in recording data, and variations in the stone's shape and mass. It is important to control these factors as much as possible to obtain accurate results.

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