How High is the Building and What is the Stone's Velocity Before Impact?

  • Thread starter libido_07
  • Start date
In summary, a stone is dropped from the top of a building and takes 5 seconds to hit the ground. Using the equations d=Vit + 1/2[g(t^2)] and Vf=Vi + gt, the height of the building can be calculated to be 122.5m downward and the velocity just before impact is 49m/s downward. However, it is unclear where the initial velocity of 1m/s came from, and it is possible that it was not given in the problem.
  • #1
libido_07
10
0

Homework Statement


A stone is dropped from the top of a building. If it takes 5 seconds for the stone to hit the ground, how high is the building? What is the velocity just before it hits?


Homework Equations


d=Vit + 1/2[g(t^2)]
Vf=Vi + gt

The Attempt at a Solution


=1m/s(5s) + 1/2[9.8m/s^2(5s^2)]
=122.5m downward
=0 + 9.8m/s^2(5s)
=49m/s downward
 
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  • #2
Where did you get the initial velocity from?
 
  • #3
that is the problem...cause that solution was given to us by our teacher...and I'm wondering how did she get it...
 
  • #4
If the stone is dropped that assumes that the initial velocity is 0. If it was stated the stone was thrown with an initial velocity of 1m/s then the solution would make sense.

From what you have put though the answer is correct and the teacher hasn't included the initial velocity to get the height of the building in the calculation. Perhaps just an error on her part, teachers are humans too you know. :wink:
 
  • #5
tnx...
-stacy libido
 

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