A falling stone takes Δt = 0.25 s to travel past a window 2.2 m tall
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falling stone takes 0.28 s to travel past a window 2.2 m tall. From what height above the top of the window did the stone fall? So from this problem I think I have these variables: vo = ? (or 0 because falling usually indicates 0 for an initial velocity?) yo = 0 y = 2.2m t = 0.28s a = -9.8m/s^2 Homework Equations Kinematic equations: y=yo+vo*t + 1/2*a*t^2 v^2 = v^20 + 2a(y-y0) The Attempt at a Solution [/B] I originally tried to use the first equation above, and solve for vo if I assumed it was not equal to zero. So I isolated vo and got, vo = -2.2 + 1/2(-9.8m/s^2)(0.28s)^2 / -0.28s vo = 9.229m/s After which I tried to substitute that value in my second equation which also didn't give me the correct answer for height (which i think is the new yo) My textbook gives me the final answer but not the steps that it reached to get to it. I have included the final answer here in white ink (so highlight to read) In case anyone wants to check their work or work backwards: [COLOR=#black]2.1m [/COLOR] I feel like my process and identifying my variables given correctly is wrong and I'm at that point where I am confusing myself repeating this question. Guidance would greatly be appreciated, Thank you all! Newbie in Physics
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