A Falling Stone

[jena]

Hi,
I have a question about height and a falling stone. The question reads:

Question: A stone falling takes 0.28 seconds to travel past a window 2.2 meters tall. From what height above the tope of the window did the stone fall?

Answer: 3.46 meters

Is this the correct answer.

Thank You

[dextercioby]

How did you do it?

Daniel.

[DaveC426913]

i.e. show your work if you want help.

[wintercarver]

heh, i remember doing that problem out of giancoli years ago.

[jena]

Sorry here's my work.
Work:

t=0
y(i)=0
v=0
V(i)=7.86 m/s
a= -9.8 m/s^2



y= (v^2-V(i)^2)/2a

y=(0-(7.86)^2)/(2*.9.8)

y=3.46 m

Thank You

[ramollari]

Sorry here's my work.
Work:

t=0
y(i)=0
v=0
V(i)=7.86 m/s
a= -9.8 m/s^2


What do you denote by v(i) and how did you derived it? If you mean the speed at the top of the window I assume you found it as:

v_i = \frac{h - \frac{gt^2}{2}}{t},

Its value should be 6.49 m/s instead.

Also do not consider accelearation as negative. It is easier to take the y-axis pointing downward to make all vector quantities positive.



y= (v^2-V(i)^2)/2a

y=(0-(7.86)^2)/(2*.9.8)

y=3.46 m

Thank You

Almost there, but you need to arrange a little, as v_i is actually the final velocity for the part of the trajectory from drop point to the top of the window. So,

h = \frac{v_i^2}{2g}

You should get a value of 2.15 m.

[jena]

Thank you, I see where I got confused