See
http://www.jiskha.com/display.cgi?id=1285261758
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?
3 answers
i still don't understand parts b and c.
(b)
let u = velocity at the top of the window.
In t=0.3 second, the object has travelled -2.2 m with an acceleration of g=-9.8 m/s².
Use the kinematic equation
S(distance)=ut+(1/2)gt²
Solve for u to get
u=S/t-(1/2)gt
=-2.2/0.3-(-9.8)/2*0.3
= the same expression Mr.Pursley gave.
(c)
Equate potential and kinetic energies to get
mgh=(1/2)mu²
where u is obtained from (b) above.
Solve for h.
let u = velocity at the top of the window.
In t=0.3 second, the object has travelled -2.2 m with an acceleration of g=-9.8 m/s².
Use the kinematic equation
S(distance)=ut+(1/2)gt²
Solve for u to get
u=S/t-(1/2)gt
=-2.2/0.3-(-9.8)/2*0.3
= the same expression Mr.Pursley gave.
(c)
Equate potential and kinetic energies to get
mgh=(1/2)mu²
where u is obtained from (b) above.
Solve for h.
2 answers