posted by joyce .
A stone is dropped off the science building and accelerates, from rest, toward the ground at 9.8 m/s/s. A curious physics student looks out the third floor window as the stone falls past. She happens to have a stopwatch and she finds that it takes 0.30 sec for the stone to fall past the 2.2 m tall window. She then sketches the velocity vs. time plot shown below, but realizing she is late for lunch, she doesn't use the plot to analyze the motion of the stone.
A) WHat was the average velocity of the stone as it fell past the window?
B) What was the velocity of the stone at the top of the window?
C) From what height above the top of the window did the stone start its fall?
average velocity= 2.2/.3 m/s
So, if the average velocity is that, that is the velocity at the center. What is the velocity at the top of the window?
Vf(at top)=-2.2/.3-9.8(-.15) where t is before the center, at t=-.15
Vf you calculate.
Vf(top of window)^2=2*g*d
solve for distance d.
i still don't understand how to do parts b and c.