What is the height of the ledge from which the flowerpot fell?

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The discussion focuses on calculating the height from which a flowerpot fell, using the equations of motion under gravity. The key equations identified are d = v1(t) + 1/2(a)(t²) and d = 1/2*g*t², where 'g' represents the acceleration due to gravity. The problem involves determining the height above a 4.0 m window, with the pot taking 0.20 seconds to fall past it. By setting up two equations, one for the distance fallen to the top of the window and another for the distance to the bottom, the solution can be derived.

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Having trouble with the equation to solve the problem.

A flowerpot falls from the ledge of an apartment building. A person in an apartment below, coincidentally in possession of a high-precision timing system, notices that it takes 0.20 seconds for the pot to fall past his 4.0 m high window. How far above he top of the window is the ledge from which the pot fell? ( Neglect any effects due to air resistance.)
 
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What equation relates distance time and acceleration directly?
 
d= v1(t)+1/2(a)(t) squared

d= ((v1+v2)/2) * t

is that correct?
 
Where you get v2 from?

There is only 1 thing moving. That second equation is some kind of average. Most texts have the free fall equations in them already derived. The first one you have is what you need. Now just think about what the velocity means and you can get your distance.

http://en.wikipedia.org/wiki/Free-fall
 
Susanem7389 said:
d= v1(t)+1/2(a)(t) squared

d= ((v1+v2)/2) * t

is that correct?

The first one is useful. So let's use it.

Can't you determine what the height is it fell from rest to the top of the window?

d = 1/2*g*t2

and then can't you also write another equation for when it passes the bottom of the window?

d + 4 = 1/2*g*(t + .2)2

Glory be. 2 equations and 2 unknowns. The answer can't be far away now.
 

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