I cannot answer this question for my life

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In summary, to find the distance above the window where the stone was released, use the equation d=vi*t+(1/2)a*t^2 and solve for vi. Then, find the distance a stone would have to fall to reach that speed and that will be the answer.
  • #1
High_Voltage
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Homework Statement


A falling stone takes .28s to travel past a window 2.2m tall. How far above the window was it released


Homework Equations


vf=vi+at
d=vit+(1/2)at^2
vf^2=vi^2+2ad


The Attempt at a Solution


vf^2=(6.5)+2(9.81 x 2.2)

2.2=(x)(.28s)+(1/2)(9.81)(.28^2)
 
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  • #2
Hi High_Voltage and welcome to PF. you have the right relevant equations, but you are not applying them correctly. Start with d=vi*t+(1/2)a*t^2
You know d=2.2 m, t=0.28 s and a = 9.8 m/s2 (assuming that "down" is positive). You don't know what vi is but you can find it from this equation. It represents the speed of the stone when it first appears at the top of the window. Once you have a number for vi you need to ask and answer another question: How far must a stone (that is released from rest) have to drop so that is reaches this speed vi? That's the answer you are looking for.
 

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