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The discussion centers on calculating the height from which a wrench fell from a helicopter, taking 9.0 seconds to reach the ground. The initial assumption was to use average velocity with the formula d = v/t, leading to an incorrect height of 4.9 meters. The correct approach involves using the formula y = ut + 0.5at², where the initial velocity (u) is zero and acceleration (a) is due to gravity. This calculation reveals the correct height as 400 meters. The confusion arises from misapplying the average velocity formula instead of accounting for the effects of gravity on the falling object.
McKeavey
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During construction of the Canada pavilion in Vancouver, a wrench fell out of a hovering helicopter and took 9.0 s to reach the ground.
How far was the helicopter from the ground?

I'd assume you find Vavg.
And put it into d = v/t
But when I do that I end up with d = 4.9.
The answer says it's 4.0 x 10^2 m AKA 400m.

What did I do wrong?
And there's no additional information given from the Q.
 
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You know that y=ut+0.5at2.

Assume the initial velocity is zero.

You are given time, what is the height y?
 
Oh..
What would be wrong with using d = v/t though? :o
 
McKeavey said:
Oh..
What would be wrong with using d = v/t though? :o

How you plan to go about finding 'v'?
 

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