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

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A stone dropped from a building takes 5 seconds to reach the ground, leading to calculations for the building's height and the stone's impact velocity. Using the equation d = Vit + 1/2[g(t^2)], the height is determined to be 122.5 meters, assuming an initial velocity of 0. The final velocity before impact is calculated as 49 m/s downward, based on the formula Vf = Vi + gt. There is confusion regarding the initial velocity, as the teacher's solution implies a non-zero initial velocity, which contradicts the problem's parameters. The discussion highlights the importance of clarity in problem statements and acknowledges potential errors by educators.
libido_07
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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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Where did you get the initial velocity from?
 
that is the problem...cause that solution was given to us by our teacher...and I'm wondering how did she get it...
 
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:
 
tnx...
-stacy libido
 
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