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

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A flowerpot falls from a ledge, taking 0.20 seconds to pass a 4.0 m high window. The relevant equations for calculating the distance fallen are discussed, specifically d = v1(t) + 1/2(a)(t^2) and d = (v1 + v2)/2 * t. The first equation is deemed more useful for this scenario, as it relates distance, time, and acceleration due to gravity. Two equations are established to solve for the height of the ledge above the window, leading to a solvable system of equations. The discussion emphasizes the application of free-fall physics to determine the height from which the pot fell.
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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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