A stone is observed falling past a small window 1 meter tall in 0.2 seconds. how fast is it moving as it passes the bottom of the window?how far is the bottom of the window from the roof, where the stone originally began its fall from rest?what is the stone's average speed as it passes the window?

Question

i can't figure out a way to do it, can any one please give me some suggestions????
THX A LOT!!!!

Answer 1

distance = Vot + 1/2(g)t^2
1 m = Vo(0.2) + 1/2(9.8)*(0.2)^2
solve for Vo which is the velocity at the top of the window. I have 4.02 m/s at the top of the window.
Use Vo as V for V = Vo + gt. At rest Vo = 0 and
4.02 + 9.8*t and solve for t, the time to fall from rest to the top of the window.
I have 0.41 seconds.
then distance = Vo*t + 1/2(9.8)(0.41)^2
Vo is 0 at the beginning. Check my thinking. Check my work.

Answer 2

Following your thinking,
I first got Vo=4.02 m/s = the speed at the top of the window
then I use the equation:Vf=Vo+gt,
so I got Vf=5.981m/s= the speed at the bottom of the window
and then I use the equation: Vf^2-Vo^2=2gd,
thus I got d=1.82m = the distance from the roof to the bottom of the window
at last, I use the equation: (Vf+Vo)/2,
finally I got the average speed as the stone passes the window = 5m/s

Is it correct?

Answer 3

My numbers were the same as yours.